TS 10th Class Maths Solutions Chapter 14 Statistics InText Questions

Students can practice 10th Class Maths Textbook SSC Solutions Telangana Chapter 14 Statistics InText Questions to get the best methods of solving problems.

TS 10th Class Maths Solutions Chapter 14 Statistics InText Questions

Think – Discuss

Question 1.
The mean value can be calculated from both ungrouped and grouped data. Which one do you think is more accurate? Why? (Page No. 327) (AS2. AS3)
Solution:
Mean calculated from ungrouped data is more accurate than, mean calculated from the grouped data. Since its calculation takes the observations in the data into consideration.

Question 2.
When it Is more convenient to use grouped data for analysis? (AS3) (Page No. 327)
Solution:
Grouped data is convenient when the values of fi and xi are low.

Question 3.
Is the result obtained by all the three methods the same? (AS3) (Page No. 331)
Solution:
Yes, mean obtained by all the three methods is same.

TS 10th Class Maths Solutions Chapter 14 Statistics InText Questions

Question 4.
If xi and fi are sufficiently small, then which method is an appropriate choice? (AS3) (Page No. 331)
Solution:
Direct method.

Question 5.
If xi and fi are numerically large numbers then which methods are appropriate to use? (AS3) (Page No. 331)
Solution:
Assumed mean method and step deviation method.

Do This

Question 1.
Find the mode of the following data. (AS1) (Page No. 334)
a) 5, 6, 9, 10, 6, 12, 3, 6, 11, 10, 4, 6, 7
Solution:
Mode : 6 (Most repeated value of the data)

b) 20, 3, 7, 13, 3, 4, 6, 7, 19, 15, 7, 18, 3
Solution:
Mode : 3, 7 (Most repeated values)

c) 2, 2, 2, 3, 3, 3, 4, 4, 4, 5, 5, 5, 6, 6, 6
Solution:
Mode : 2, 3, 4, 5, 6 (Most repeated values)

TS 10th Class Maths Solutions Chapter 14 Statistics InText Questions

Question 2.
Is the mode always at the center of the data ? (AS3) (Page No. 334)
Solution:
No, Mode may not beat the centre always.

Question 3.
Does the mode change ? If another observation is added to the data in Example ? Comment. (Page No. 334)
Solution:
The data given in example 4 is as follows :
0, 1, 2, 2, 2, 3, 3, 4, 5, 6.
Except 2 and 3, the other observations occur only once. 2 occurs three times while 3 occurs two times. Any other observation except 2 and 3 is added, the mode of the data will not be affected. If 3 is added, then 3 occurs three times. There are already three 2’s in the data. Hence, the data has two modes (i.e.,) 3 and 2. If 2 is added, then 2 occurs four times. So, the mode of the data will be 2, (i.e.,) the mode of the data is not affected because on observing the data prior to the addition of 2, we can say that the mode is 2.

Question 4.
If the maximum value of an observation in the data in example 4 is changed to 8, would the mode of the data be affected ? Comment. (Page No. 334)
Solution:
If the maximum value is altered to 8, the mode remains the same. Mode doesn’t consider the values but consider their frequencies only.

Think – Discuss

Question 1.
It depends upon the demand of the situation whether we are interested in finding the average marks obtained by the students or the marks obtained by most of the students. (AS3) (Page No. 336)
a) What do we find in the first situation ?
Solution:
We find A.M.

TS 10th Class Maths Solutions Chapter 14 Statistics InText Questions

b) What do we find in the second situation?
Solution:
We find the mode.

Question 2.
Can mode be calculated for grouped data which unequal class sizes ? (Mar. ’15 (AP)) (Page No. 336)
Solution:
Yes. mode can be calculated for grouped data with unequal class sizes.

TS 10th Class Maths Solutions Chapter 14 Statistics Ex 14.2

Students can practice 10th Class Maths Textbook SSC Solutions Telangana Chapter 14 Statistics Ex 14.2 to get the best methods of solving problems.

TS 10th Class Maths Solutions Chapter 14 Statistics Exercise 14.2

Question 1.
The following table shows the ages of the patients admitted in a hospital during a year :

Age (in years)5-1515-2525-3535-4545-5555-65
Number of patients6112123145

Find the mode and the mean of the data given above. Compare and interpret the two measures of central tendency. (AS3, AS4)
Solution:
Maximum number of patients joined in the age group 35 – 45
∴ Modal class is 35 – 45.
Lower limit of the modal class ‘l’ = 35
class size, h = 10
Frequency of modal class. f1 = 23
Frequency of the class preceding the modal class f0 = 21
Frequency of the class succeding the modal class f2 = 14
TS 10th Class Maths Solutions Chapter 14 Statistics Ex 14.2 1
∴ Mode = l + \(\frac{\left(f_1-f_0\right)}{2 f_1-f_0-f_2}\) × h
= 35 + \(\left[\frac{23-21}{2 \times 23-21-14}\right]\) × 10
= 35 + \(\left[\frac{2}{46-35}\right]\) × 10
= 35 + \(\frac{2}{11}\) × 10 = 35 + 1.81818 ….
= 36.8 years

Mean \(\overline{\mathrm{x}}\) = \(\frac{\Sigma \mathrm{f}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}}{\Sigma \mathrm{f}_{\mathrm{i}}}\) = \(\frac{2830}{80}\) = 35.37 years.

Interpretation : Mode age is 36.8 years, Mean age = 35.37 years.
Maximum number of patients admitted in the hospital are the age 36.8 years, while on an average the age of patients admitted to the hospital is 35.37 years. Mode is less than the mean.

Question 2.
The following data gives the information on the observed life times (in hours) of 225 electrical components : (AS4)

Lifetimes (in hours)0 – 2020 – 4040 – 6060 – 8080 – 100100 – 120
Frequency103552613829

Determine the modal life times of the components.
Solution:

Class intervalFrequencies
0-2010
20-4035
40-6052
60-8061
80 – 10038
100 – 12029

Since the maximum frequency 61 is in the class 60 – 80, this is the required modal class.
Modal class frequency, f1 = 61
Frequency of the class preceding the modal class f0 = 52
Frequency of the class succeding the modal class f2 = 38
lower boundary of the modal class l = 60
Height of the class, h = 20
∴ Mode (z) = l + \(\frac{\left(f_1-f_0\right)}{2 f_1-\left(f_0+f_2\right)}\) × h
= 60 + \(\left[\frac{61-52}{2 \times 61-(52+38)}\right]\) × 20
= 60 + \(\frac{9}{122-90}\) × 20
= 60 + \(\frac{9}{32}\) × 20
= 60 + 5.625 = 65.625 hours.

TS 10th Class Maths Solutions Chapter 14 Statistics Ex 14.2

Question 3.
The following data gives the distribution of total monthly household expenditure of 200 families of Gummadidala village. Find the modal monthly expenditure of the families. Also find the mean monthly expenditure. (AS4)

Expenditure (in rupees)1000-15001500-20002000-25002500-30003000-35003500-40004000-45004500-5000
Number of families244033283022167

Solution:
TS 10th Class Maths Solutions Chapter 14 Statistics Ex 14.2 2
Assumed mean (a) = 3250
Σfi = 200; Σfiui = – 235
Mean monthly income = \(\overline{\mathrm{x}}\) = a + \(\frac{\Sigma \mathrm{f}_{\mathrm{i}} \mathrm{u}_{\mathrm{i}}}{\Sigma \mathrm{f}_{\mathrm{i}}}\) × h
= 3250 – \(\frac{235}{200}\) × 500
= 3250 – 587.5
= ₹ 2662.50
Since the maximum families 40 lies in the class 1500 – 2000, this is the required modal class.
Lower boundary of modal class (l) = 1500
Frequency of the modal class (f1) = 40
Frequency of the class preceding the modal class f0 = 24
Frequency of the class succeding the modal class f2 = 33
Height of the class, h = 500
Mode (z) = l + \(\frac{\left(f_1-f_0\right)}{2 \times f_1-\left(f_0+f_2\right)}\) × h
⇒ 1500 + \(\frac{40-24}{2 \times 40-(24+33)}\) × 500
= 1500 + \(\frac{16 \times 500}{80-57}\) = 1500 + \(\frac{8000}{23}\)
⇒ 1500 + 347.826
= ₹ 1847.83
Hence, modal monthly income = ₹ 1847.83

TS 10th Class Maths Solutions Chapter 14 Statistics Ex 14.2

Question 4.
The following distribution gives the state-wise, teachers, student ratio in higher second’ ary schools of India. Find the mode and mean of this data. Interpret the two measures. (AS3, AS4)

Number of students15-2020-2525-3030-3535-4040-4545-5050-55
Number of States389103002

Solution:
TS 10th Class Maths Solutions Chapter 14 Statistics Ex 14.2 3
Mean (\(\overline{\mathrm{x}}\)) = a + \(\frac{\Sigma \mathrm{f}_{\mathrm{i}} \mathrm{u}_{\mathrm{i}}}{\Sigma \mathrm{f}_{\mathrm{i}}}\) × h
a – assumed mean = 32.5; h – height of the class = 5
∴ \(\overline{\mathrm{x}}\) = 32.5 – \(\frac{22}{35}\) × 5 = 32.5 – 3.28 = 29.22
Since the maximum number of states ‘10′ lies in the class interval 30 – 35, this is the modal class.
Lower boundary of the modal class, l = 30
Frequency of the modal class, f1 = 10
Frequency of the class preceding the modal class, f0 = 9
Frequency of the class succeding the modal class, f2 = 3
Height of the class h = 5
Mode (z) = l + \(\left[\frac{f_1-f_0}{\left(f_1-f_0\right)+\left(f_1-f_2\right)}\right]\) × h
⇒ 30 + \(\frac{10-9}{(10-9)+(10-3)}\) × 5
= 30 + \(\frac{1 \times 5}{1+7}\) = 30 + \(\frac{5}{8}\)
= 30 + 0625 ⇒ 30.625.
Mode of states have a teacher students ratio 30.625 and on an average of this ratio is 29.22.

Question 5.
The given distribution shows the number of runs scored by some top batsmen of the world in one international cricket matches.

Runs3000-40004000-50005000-60006000-70007000-80008000-90009000-1000010000-11000
Number of batsmen418976311

Find the mode of the data. (AS4)
Solution:

Class intervalFrequencies
3000 – 40004
4000 – 500018
5000 – 60009
6000 – 70007
7000 – 80006
8000 – 90003
9000 – 100001
10000- 110001

Maximum no.of batsmen are in the class 4000 – 5000
Modal class is 4000 – 5000
Lower boundary of the modal class ‘l’ = 4000
frequency of the modal class, f1 = 18
frequency of the class preceding the modal class f0 = 4
frequency of the class succeeding the modal class f2 = 9
Size of the class, h = 1000
Mode(z) = l + \(\left[\frac{\mathrm{f}_1-\mathrm{f}_0}{\left(\mathrm{f}_1-\mathrm{f}_0\right)+\left(\mathrm{f}_1-\mathrm{f}_2\right)}\right]\) × h
Mode (z) = 4000 + \(\frac{18-4}{(18-4)+(18-9)}\) × 1000
= 4000 + \(\frac{14}{14+9}\) × 1000
= 4000 + \(\frac{14000}{23}\) = 4000 + 608.695
= 4608.69
≅ 4608.7 runs

TS 10th Class Maths Solutions Chapter 14 Statistics Ex 14.2

Question 6.
A student noted the number of cars passing through a spot on a road for 100 periods, each of 3 minutes and summarised this in the table given below.

Number of cars0 – 1010 – 2020 – 3030 – 4040 – 5050 – 6060 – 7070 – 80
Frequency71413122011158

Find the mode of the data. (AS4)
Solution:

No.of carsFrequency
0 – 107
10 – 2014
20 – 3013
30 – 4012
40 – 5020
50 – 6011
60 – 7015
70 – 808

Since, the maximum frequency is 20, the modal class is 40 – 50.
Lower boundary of the modal class, ‘l’ = 40
Frequency of the modal class, f1 = 20
Frequency of the class preceding the modal class f0 = 12
Frequency of the class succeeding the modal class f2 = 11
Height of the class, h = 10;
Mode (z) = l + \(\left[\frac{\mathrm{f}_1-\mathrm{f}_0}{\left(\mathrm{f}_1-\mathrm{f}_0\right)+\left(\mathrm{f}_1-\mathrm{f}_2\right)}\right]\) × h
= 40 + \(\frac{(20-12)}{(20-12)+(20-11)}\) × 10
= 40 + \(\frac{8}{8+9}\) × 10
= 40 + \(\frac{80}{17}\)
= 40 + 4.70588
= 44.705
≅ 44.7 cars.

TS 10th Class Maths Solutions Chapter 14 Statistics Ex 14.4

Students can practice 10th Class Maths Textbook SSC Solutions Telangana Chapter 14 Statistics Ex 14.4 to get the best methods of solving problems.

TS 10th Class Maths Solutions Chapter 14 Statistics Exercise 14.4

Question 1.
The following distribution gives the daily income of 50 workers of a factory.

Daily income (in Rupees)250-300300-350350-400400-450450-500
Number of workers12148610

Convert the distribution above to a less than type cumulative frequency distribution and draw its ogive. (AS5)
Solution:
For less than ogive, we take the upper class limits on X-axis and their corresponding cumulative frequencies on Y – axis, choosing a convenient scale.

Daily income (in Rupees)250-300300-350350-400400-450450-500
Number of workers12148610
Class Interval (Upper Limits)frequencycumulative frequencyPoints
3001212(300, 12)
3501426(350, 26)
400834(400, 34)
450640(450, 40)
5001050(500, 50)

The points to be plotted are (300, 12) (350, 26) (400, 34) (450, 40) and (500, 50)
Scale :
On X-axis : 1 cm = 50 units
On Y-axis : 1 cm = 5 units
TS 10th Class Maths Solutions Chapter 14 Statistics Ex 14.4 1

Question 2.
During the medical check-up of 35 students of a class, their weights were recorded as follows : (AS5)

Weight (in kg)Number of students
Less than 380
Less than 403
Less than 425
Less than 449
Less than 4614
Less than 4828
Less than 5032
Less than 5235

Draw a less than type ogive for the given data. Hence obtain the median weight from the graph and verify the result by using the formula.
Solution:
The points to be plotted on a graph paper are : (38, 0), (40, 3), (42, 5), (44, 9), (46, 14), (48, 28), (50, 32), (52, 35)
Scale :
On X-axis : 1 cm = 2 units
On Y-axis : 1 cm = 2 units
TS 10th Class Maths Solutions Chapter 14 Statistics Ex 14.4 2
Number of observations = 35
Here \(\frac{\mathrm{n}}{2}\) = \(\frac{35}{2}\) = 17.5
Locate the point 17.5 on the Y – axis. From the point, draw a line parallel to the X – axis cutting the curve at a point. From the point, draw a perpendicular to the X – axis.

TS 10th Class Maths Solutions Chapter 14 Statistics Ex 14.4

The point of intersection of this perpendicular with the X – axis determines the median of the given data as 46.8 kg.

WeightNumber of Students (c.f)Frequency (f)
Below 3800
38-4033
40-4252
42-4494
44-46145
46-482814
18-50324
50-52353

Number of observations = n = 35
\(\frac{\mathrm{n}}{2}\) = \(\frac{35}{2}\) = 17.5
17.5 belongs to the class 46 – 48
∴ Median class = 46 – 48
l – lower boundary of class = 46
f – frequency of the median class = 14
c.f. = 14
Class size = 2
Median = l + \(\frac{\left(\frac{n}{2}-c . f .\right)}{f}\) × h
= 46 + \(\frac{17.5-14}{14}\) × 2
= 46 + \(\frac{3.5}{14}\) × 2
= 46 + \(\frac{7}{14}\) = 46 + \(\frac{1}{2}\) = 46.5
Hence, median is 46.5 either ways.

TS 10th Class Maths Solutions Chapter 14 Statistics Ex 14.4

Question 3.
The following table gives production yield per hectare of wheat of 100 farms of a village. (AS5)

Production yield (Qui/Hec)50-5555-6060-6565-7070-7575-80
Number of farmers2812243816

Change the distribution to a more than type distribution and draw its ogive.
Solution:
More than type distribution
TS 10th Class Maths Solutions Chapter 14 Statistics Ex 14.4 3
Points to be plotted on the graph paper are:
(50, 100), (55, 98), (60, 90), (65. 78), (70, 54) and (75, 16)
Scale:
On X-axis: 1 cm = 5 units
On Y-axis: 1 cm = 10 units
TS 10th Class Maths Solutions Chapter 14 Statistics Ex 14.4 4

TS 10th Class Maths Solutions Chapter 14 Statistics Ex 14.3

Students can practice 10th Class Maths Textbook SSC Solutions Telangana Chapter 14 Statistics Ex 14.3 to get the best methods of solving problems.

TS 10th Class Maths Solutions Chapter 14 Statistics Exercise 14.3

Question 1.
The following frequency distribution gives the monthly consumption of electricity of 68 consumers of a locality. Find the median, mean and mode of the data and compare them. (AS4)

Monthly consumption65-8585-105105-125125-145145-165165-185185-205
Number of consumers4513201484

Median :
Solution:
TS 10th Class Maths Solutions Chapter 14 Statistics Ex 14.3 5
Here n = 68; \(\frac{\mathrm{n}}{2}\) = \(\frac{68}{2}\) = 34
The median lies in the class 125 – 145.
Lower limit (l) of the median class = 125
Frequency of the median class (f) = 20 of (cumulative frequency of the class 105 – 125) = 22
class size (h) = 20
∴ Median = l + \(\frac{\left[\frac{\mathrm{n}}{2}-\mathrm{c.f}\right]}{\mathrm{f}}\) × h
= 125 + \(\frac{\left[\frac{n}{2}-c f\right]}{f}\) × 20
= 125 + 12
= 137 units.

Mean :
TS 10th Class Maths Solutions Chapter 14 Statistics Ex 14.3 1
Mean (\(\overline{\mathrm{X}}\)) = a + \(\frac{\Sigma \mathrm{f}_{\mathrm{i}} \mathrm{u}_{\mathrm{i}}}{\Sigma \mathrm{f}_{\mathrm{i}}}\) × h
Here a = 135, h = 20, Σfi = 68, Σfiui = 7
= 135 + \(\left[\frac{7}{68}\right]\) × 20
= 135 + \(\frac{35}{17}\)
= 135 + 2.05
= 137.05

Mode :
Since the maximum number of consumers have their monthly consumption (in units) in the interval 125 – 145
∴ lower limit of the modal class (l) = 125
class size (h) = 20
Frequency of the modal class (f1) = 20
Frequency of the class preceding the modal class (f0) = 13
Frequency of the class succeeding the modal class (f2) = 14
Modal = l + \(\frac{\left(f_1-f_0\right)}{\left(2 f_1-f_0-f_2\right)}\) × h
= 125 + \(\frac{(20-13)}{(2 \times 20-13-14)}\) × 20
= 125 + \(\frac{7}{(40-13-14)}\) × 20
= 125 + \(\frac{7 \times 20}{13}\) = 125 + \(\frac{140}{13}\)
= 125 + 10.76 = 135.76 units

Comparison :
In this case, the three measures i.e., mean, median and mode are approximately equal.

TS 10th Class Maths Solutions Chapter 14 Statistics Ex 14.3

Question 2.
If the median of 60 observations, give below is 28.5, find the value of x and y. (AS1)

Class interval0-1010-2020-3030-4040-5050-60
Frequency5x2015y5

Solution:
TS 10th Class Maths Solutions Chapter 14 Statistics Ex 14.3 6
Hence, n = 60 (given)
\(\frac{\mathrm{n}}{2}\) = \(\frac{60}{2}\) = 30
⇒ 45 + x + y = 60
⇒ x + y = 60 – 45 = 15 ……………… (1)
The median is given as 28.5.
If lies in the class 20 – 30.
So, l = 20;
Frequency of the median class (f) = 20
cf (cumulative frequency of the class preceding the median class 10 – 20) = 5 + x
∴ Median = l + \(\frac{\left[\frac{\mathrm{n}}{2}-\mathrm{c.f}\right]}{\mathrm{f}}\) × h
⇒ 28.5 = 20 + \(\left[\frac{30-(5+x)}{20} \times 10\right]\)
⇒ 28.5 = 20 + \(\frac{25-x}{2}\)
⇒ \(\frac{25-x}{2}\) = 28.5 – 20 = 8.5
⇒ 25 – x = 2 × 8.5
⇒ 25 – x = 17
⇒ x = 25 – 17 = 8 ……………… (2)
from (1) and (2) : we get
8 in x + y = 15, 8 + y = 15
∴ y = 15 – 8 = 7
Hence, x = 8 and y = 7

TS 10th Class Maths Solutions Chapter 14 Statistics Ex 14.3

Question 3.
A life insurance agent found the following data about distribution of ages of 100 policy holders. Calculate the median age. (policies are given only to persons having age 18 years onwards but less than 60 years.) (AS4)

Age (in years)Below 20Below 25Below 30Below 35Below 40Below 45Below 50Below 55Below 60
Number of policy holders26244578899298100

We find the class intervals and their corresponding frequencies to calculate the median age.
Solution:
TS 10th Class Maths Solutions Chapter 14 Statistics Ex 14.3 2
Hence, n = 100: \(\frac{\mathrm{n}}{2}\) = \(\frac{100}{2}\) = 50
The median lies in the class 35 – 40. So l = 35
frequency of the median class (f) = 33
cf (cumulative frequency of the class preceeding the median dass 30 – 35) = 45
class size(h) = 5
∴ Median = l + \(\frac{\left[\frac{\mathrm{n}}{2}-\mathrm{c.f}\right]}{\mathrm{f}}\) × h
= 35 + \(\frac{(50-45)}{33}\) × 5
= 35 + \(\frac{5 \times 5}{33}\) = 35 + \(\frac{25}{33}\)
= 35 + 0.76 = 35.76
Hence, the median age = 35.76 years.

Question 4.
The lengths of 40 leaves of a plant are measured correct to the nearest millimetre and the data obtained Is represented in the following table: (AS4)

Length (in mm)118-126127-135136-144145-153154-162163-171172-180
Number of leaves35912542

Find the median length of the leaves (Hint : The data needs to be converted to continuous classes for finding the median, since the formula assumes continuous classes. The classes then change to 117.5 – 126.5, 126.5 – 135.5 ………. 171.5 – 180.5)
Solution:
TS 10th Class Maths Solutions Chapter 14 Statistics Ex 14.3 3
Here, n = 40; \(\frac{\mathrm{n}}{2}\) = \(\frac{40}{2}\) = 20
The median class is 144.5 – 153.5
Lower limit (l) of the median class = 144.5
cf (cumulative frequency of class preceding the median class 144.5 – 153.5) = 17
f (frequency of the median class) = 12
h (class size) = 9
using the formula, median
= l + \(\frac{\left[\frac{\mathrm{n}}{2}-\mathrm{c.f}\right]}{\mathrm{f}}\) × h
= 144.5 + \(\frac{(20-17) \times 9}{12}\) × 9
= 144.5 + \(\frac{3 \times 9}{12}\) = 144.5 + \(\frac{9}{4}\)
= 144.5 + 2.25
= 146.75
∴ Median length = 146.75 mm.

TS 10th Class Maths Solutions Chapter 14 Statistics Ex 14.3

Question 5.
The following table gives the distribution of the life-time of 400 neon lamps

Life time (in hours)1500-20002000-25002500-30003000-35003500-40004000-45004500-5000
Number of lamps14566086746248

Find the median life time of a lamp. (AS4)
Solution:

Life time (in hours) class intervals (C.I)No.of lamps (f)cumulative frequency (cf)
1500 – 20001414
2000 – 25005670
2500 – 300060130
3000 – 350086216
3500 – 400074290
4000 – 450062352
4500 – 500048400

Here, n = 400; \(\frac{\mathrm{n}}{2}\) = \(\frac{400}{2}\) = 200
The median class is 3000 – 3500
lower limit (1) of the median class = 3000
cf (cumulative frequency of the class preceding the median class 2500 – 3000) = 130
f (frequency of the median class) = 86
h(class size) = 500
using the formula, Median = l + \(\frac{\left[\frac{\mathrm{n}}{2}-\mathrm{c.f}\right]}{\mathrm{f}}\) × h
= 3000 + \(\frac{(200-130) \times 500}{86}\)
= 3000 + \(\frac{70 \times 500}{86}\)
= 3000 + 406.98
= 3406.98 hours
∴ Median life = 3406.98 hours.

TS 10th Class Maths Solutions Chapter 14 Statistics Ex 14.3

Question 6.
loo surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters ¡n the English alphabet in the surnames was obtained as follows:

Number of letters1-44-77-1010-1313-1616-19
Number of surnames630401644

Determine the median number of letters In the surnames. Find the mean number of letters in the surnames P Also, find the modal size of the surnames. (AS4)
Solution:
Median:

No.of letters class intervals (C.I)No.of surnames (f)cumulative frequency (cf)
1-466
4-73036
7-104076
10-131692
13-16496
16-194100

Here, n = 100; \(\frac{\mathrm{n}}{2}\) = \(\frac{100}{2}\) = 50
So, the median lies in the class 7 – 10
lower limit (1) of the median class = 7
cf (cumulative frequency of the class preceding median class 7 – 10) = 36
f (frequency of the median class) = 40
h (class size) = 3
using the formula. Median
= l + \(\frac{\left[\frac{\mathrm{n}}{2}-\mathrm{c.f}\right]}{\mathrm{f}}\) × h × h
= 7 + \(\frac{(50-36) \times 3}{40}\)
= 7 + \(\frac{14 \times 3}{40}\) = 7 + \(\frac{21}{20}\)
= 7 + 1.05 = 8.05
Hence, the median of letters in the surnames = 8.05

Mean :
TS 10th Class Maths Solutions Chapter 14 Statistics Ex 14.3 4
Mean (\(\overline{\mathrm{x}}\)) = a + \(\left[\frac{\Sigma \mathrm{f}_{\mathrm{i}} \mathrm{u}_{\mathrm{i}}}{\Sigma \mathrm{f}_{\mathrm{i}}} \times \mathrm{h}\right]\)
Here a = 8.5, h = 3, Σfi = 100, Σfiui = -6
= 8.5 – \(\frac{6 \times 3}{100}\)
= 8.5 – \(\frac{18}{100}\)
= 8.5 – 0.18
= 8.32
Hence, the mean of the surnames = 8.32

Mode:

Class intervals (C.I)Frequency (f)
1-46
4-730
7-1040
10-1316
13-164
16 -194

Since, the maximum number of surnames have number of letters in the interval 7 – 10, the modal class is 7 – 10
Lower limit (1) of the modal class = 7
frequency of the modal class (f1) = 40
Frequency of the class preceding the modal class (f0) = 30
frequency of the class succeeding the modal class (f2) = 16
class size(h) = 3
∴ Mode = l + \(\left[\frac{\mathrm{f}_1-\mathrm{f}_0}{2 \mathrm{f}_1-\mathrm{f}_0-\mathrm{f}_2}\right]\) × h
= 7 + \(\frac{(40-30)}{(2 \times 40-30-16)}\) × 3
= 7 + \(\left[\frac{10 \times 3}{80-30-16}\right]\)
= 7 + \(\frac{30}{34}\)
= 7 + \(\frac{15}{17}\)
= 7 + 0.88 = 7.88
Hence, the modal size of the surnames = 7.88

TS 10th Class Maths Solutions Chapter 14 Statistics Ex 14.3

Question 7.
The distribution below gives the weights of 30 students of a class. Find the median weight of the students. (AS4) (Mar ’16 (A.P))

Weight (in kg)40-4545-5050-5555-6060-6565-7070-75
Number of students2386632

Solution:

Class Intervals weight in kgs (C.I)No.of students (f)cumulative frequency (cf)
40-4522
45-5035
50 -55813
55-60619
60-65625
65-70328
70-75230

Hence, n = 30; \(\frac{\mathrm{n}}{2}\) = \(\frac{30}{2}\) = 15
So, the median lies in the class 55 – 60
∴ l = 55
frequency of the median class (f) = 6
cf (cumulative frequency, of the class 50 – 55) = 13
class size (h) = 5
∴ Median = l + \(\frac{\left[\frac{\mathrm{n}}{2}-\mathrm{c.f}\right]}{\mathrm{f}}\) × h
= 55 + \(\frac{(15-13) \times 5}{6}\)
= 55 + \(\frac{2 \times 5}{6}\)
= 55 + \(\frac{5}{3}\) = 55 + 1.67
= 56.67
Hence, the median weight of the students = 56.67 kgs.

TS 10th Class Maths Solutions Chapter 14 Statistics Ex 14.1

Students can practice 10th Class Maths Textbook SSC Solutions Telangana Chapter 14 Statistics Ex 14.1 to get the best methods of solving problems.

TS 10th Class Maths Solutions Chapter 14 Statistics Exercise 14.1

Question 1.
A survey was conducted by a group of students as a part of their environment awareness programme, in which they collected the following data regarding the number of plants in 20 houses in a locality. Find the mean number of plants per house.

Number of plants0 – 22 – 44-66 – 88 – 1010 – 1212-14
Number of houses1215623

Solution:
TS 10th Class Maths Solutions Chapter 14 Statistics Ex 14.1 1
∴ \(\overline{\mathrm{x}}\) = \(\frac{\Sigma \mathrm{f}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}}{\Sigma \mathrm{f}_{\mathrm{i}}}\) = \(\frac{162}{20}\) = 8.1
!! Since fi and xi are of small values we use direct method.

Question 2.
Consider the following distribution of daily wages of 50 workers of a factory. (AS4)

Daily wages in Rupees200 – 250250 – 300300 – 350350 – 400400- 450
Number of workers12148610

Find the mean daily wages of the workers of the factory by using an appropriate method.
Solution:
TS 10th Class Maths Solutions Chapter 14 Statistics Ex 14.1 2
Here, the xi are of large numerical values in 250 – 300. So, a = 275.
So we use Assumed Mean method then,
\(\overline{\mathrm{x}}\) = a + \(\frac{\Sigma \mathrm{f}_{\mathrm{i}} \mathrm{d}_{\mathrm{i}}}{\Sigma \mathrm{f}_{\mathrm{i}}}\)
Here, the Assumed mean is taken as 275.
∴ \(\overline{\mathrm{x}}\) = a + \(\frac{\Sigma \mathrm{f}_{\mathrm{i}} \mathrm{d}_{\mathrm{i}}}{\Sigma \mathrm{f}_{\mathrm{i}}}\) = 275 + \(\frac{1900}{50}\) = 275 + 38 = 313

TS 10th Class Maths Solutions Chapter 14 Statistics Ex 14.1

Question 3.
The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is ₹ 18. Find the missing frequency f. (AS4)

Daily pocket allowance(in Rupees)11 – 1313 – 1515 – 1717 – 1919 – 2121 – 2323 – 25
Number of children76913f54

Solution:
TS 10th Class Maths Solutions Chapter 14 Statistics Ex 14.1 3
\(\overline{\mathrm{x}}\) = \(\frac{\Sigma \mathrm{f}_{\mathrm{i}} \mathrm{d}_{\mathrm{i}}}{\Sigma \mathrm{f}_{\mathrm{i}}}\)
\(\overline{\mathrm{x}}\) = 18 (given)
⇒ 18 = \(\frac{752+20 f}{(44+f)}\)
⇒ 18(44 + f) = 752 + 20f ⇒ 20f – 18f = 792 – 752
⇒ 2f = 40
∴ f = \(\frac{40}{2}\) = 20

TS 10th Class Maths Solutions Chapter 14 Statistics Ex 14.1

Question 4.
Thirty women were examined in a hospital by a doctor and their of heart beats per minute were recorded and summarised as shown. Find the mean heart beats per minute for these women, choosing a suitable method. (AS4)

Number of heart beats/minute65-6868-7171-7474-7777-8080-8383-86
Number of women2438742

Solution:
TS 10th Class Maths Solutions Chapter 14 Statistics Ex 14.1 4
\(\overline{\mathrm{x}}\) = a + \(\frac{\Sigma \mathrm{f}_{\mathrm{i}} \mathrm{d}_{\mathrm{i}}}{\Sigma \mathrm{f}_{\mathrm{i}}}\)
= 75.5 + \(\frac{12}{30}\) = 75.5 + 0.4 = 75.9

Question 5.
In a retail market, fruit vendors were selling oranges kept in packing baskets. These baskets contained varying number of oranges. The following was the distribution of oranges. (AS4)

Number of oranges10-1415-1920-2425-2930-34
Number of baskets1511013511525

Find the mean number of oranges kept in each basket. Which method of finding the mean did you choose ?
Solution:
TS 10th Class Maths Solutions Chapter 14 Statistics Ex 14.1 5
Here, we use step division method where a = 22, h = 5
\(\overline{\mathrm{x}}\) = a + \(\left[\frac{\Sigma \mathrm{f}_{\mathrm{i}} \mathrm{u}_{\mathrm{i}}}{\Sigma \mathrm{f}_{\mathrm{i}}}\right]\) × h
⇒ 22 + \(\frac{25}{400}\) × 5
⇒ 22 + 0.31 = 22.31

TS 10th Class Maths Solutions Chapter 14 Statistics Ex 14.1

Question 6.
The table below shows the daily expenditure on food of 25 households in a locality.

Daily expenditure (in Rupees)100-150150-200200-250250-300300-350
Number of house holds451222

Find the mean daily expenditure on food by a suitable method. (AS4)
Solution:
TS 10th Class Maths Solutions Chapter 14 Statistics Ex 14.1 6
Here, a = 225, h = 50
\(\overline{\mathrm{x}}\) = a + \(\left[\frac{\Sigma \mathrm{f}_{\mathrm{i}} \mathrm{u}_{\mathrm{i}}}{\Sigma \mathrm{f}_{\mathrm{i}}}\right]\) × h
= 225 + \(\frac{(-7)}{25}\) × 50 = 225 – 14 = 211
The average daily expenditure on food = ₹ 211

TS 10th Class Maths Solutions Chapter 14 Statistics Ex 14.1

Question 7.
To find out the concentration of SO2 in the air (in parts per million, i.e., ppm), the data was collect for 30 localities in a certain city and is presented below.

Concentration of SO2 in ppm0.00-0.040.04-0.080.08-0.120.12-0.160.16-0.200.20-0.24
Frequency499242

Find the mean concentration of SO2 in the air. (AS4)
Solution:
TS 10th Class Maths Solutions Chapter 14 Statistics Ex 14.1 7
∴ \(\overline{\mathrm{x}}\) = \(\frac{\Sigma \mathrm{f}_{\mathrm{i}} \mathrm{d}_{\mathrm{i}}}{\Sigma \mathrm{f}_{\mathrm{i}}}\)
= \(\frac{2.96}{30}\) = 0.0986666 …….
≅ 0.099 ppm

Question 8.
A class teacher has the following attendence record of 40 students of a class for the whole term. Find the mean number of days a student was present out of 56 days in the term. (AS4)

Number of days35-3838-4141-4444-4747-5050-5353-56
Number of students134471011

Solution:
TS 10th Class Maths Solutions Chapter 14 Statistics Ex 14.1 8
Here, a = 51.5
∴ \(\overline{\mathrm{x}}\) = a + \(\frac{\Sigma \mathrm{f}_{\mathrm{i}} \mathrm{d}_{\mathrm{i}}}{\Sigma \mathrm{f}_{\mathrm{i}}}\)
= 51.5 – \(\frac{99}{40}\)
⇒ 51.5 – 2.475 = 49.025 ≅ 49 days

TS 10th Class Maths Solutions Chapter 14 Statistics Ex 14.1

Question 9.
The following table gives the literacy rate (in percentage) of 35 cities. Find the mean literacy rate. (AS4)

Literacy rate in %45-5555-6565-7575-8585-95
Number of cities3101183

Solution:
TS 10th Class Maths Solutions Chapter 14 Statistics Ex 14.1 9
Here a = 70, h = 10
∴ \(\overline{\mathrm{x}}\) = a + \(\frac{\Sigma \mathrm{f}_{\mathrm{i}} \mathrm{d}_{\mathrm{i}}}{\Sigma \mathrm{f}_{\mathrm{i}}}\) × h
⇒ \(\overline{\mathrm{x}}\) = 70 – \(\frac{2}{35}\) × 10
= 70 – \(\frac{20}{35}\) = 70 – 0.5714
= 69.4285 ≅ 69.43%

TS 6th Class Maths Solutions Chapter 12 Symmetry Ex 12.2

Students can practice Telangana SCERT Class 6 Maths Solutions Chapter 12 Symmetry Ex 12.2 to get the best methods of solving problems.

TS 6th Class Maths Solutions Chapter 12 Symmetry Exercise 12.2

Question 1.
Write any five man made things which have two lines of symmetry.
Answer:
Blackboard, plain door of a house, the ruleif, writing pad, towel, sponze duster.

Question 2.
Write any five natural objects which have two or more than two lines of symmetry.
Answer:
Pumpkin, watermelon, apple, guava, boiled egg.

Question 3.
Find the number of lines of symmetry for the following shapes.
TS 6th Class Maths Solutions Chapter 12 Symmetry Ex 12.2 1
Answer:
Fig (i) has 4 lines of symmetry.
Fig (ii) has 1 line of symmetry.
Fig (iii) has 2 lines of symmetry.
Fig (iv) has ‘O’ lines of symmetry.
Fig (v) has 4 lines of symmetry.
Fig (vi) has 2 lines of symmetry.
TS 6th Class Maths Solutions Chapter 12 Symmetry Ex 12.2 2

Question 4.
Draw the possible number of lines of symmetry.
TS 6th Class Maths Solutions Chapter 12 Symmetry Ex 12.2 3
Answer:
TS 6th Class Maths Solutions Chapter 12 Symmetry Ex 12.2 4

TS 6th Class Maths Solutions Chapter 12 Symmetry Ex 12.2

Question 5.
From the above problem complete the following table.

ShapeNumber of lines of symmetry
i)Equilateral triangle
ii)Isosceles triangle
iii)Scalene triangle
iv)Rhombus
v)Hexagon
vi)Circle

Answer:

ShapeNumber of lines of symmetry
i)Equilateral triangle3
ii)Isosceles triangle1
iii)Scalene triangleNo line of symmetry
iv)Rhombus2
v)Hexagon6
vi)CircleCountless (Infinite)

Question 6.
A few folded sheets and designs drawn about the fold are given. In each case, draw a rough diagram of the complete figure that would be seen when the design is cut off.
TS 6th Class Maths Solutions Chapter 12 Symmetry Ex 12.2 5
Answer:
TS 6th Class Maths Solutions Chapter 12 Symmetry Ex 12.2 6

TS 6th Class Maths Solutions Chapter 11 Ratio and Proportion InText Questions

Students can practice Telangana SCERT Class 6 Maths Solutions Chapter 11 Ratio and Proportion InText Questions to get the best methods of solving problems.

TS 6th Class Maths Solutions Chapter 11 Ratio and Proportion InText Questions

Try These

Question 1.
Observe the example and fill in the blanks
TS 6th Class Maths Solutions Chapter 11 Ratio and Proportion InText Questions 1
TS 6th Class Maths Solutions Chapter 11 Ratio and Proportion InText Questions 2
Answer:
TS 6th Class Maths Solutions Chapter 11 Ratio and Proportion InText Questions 3

TS 6th Class Maths Solutions Chapter 11 Ratio and Proportion InText Questions

Try These

Question 1.
In the given figure, find the ratio of
i) Shaded parts to unshaded parts.
ii) Shaded parts to total parts.
iii) Unshaded parts to total parts.
TS 6th Class Maths Solutions Chapter 11 Ratio and Proportion InText Questions 4
Answer:
i) Shaded parts to unshaded parts = 4: 1
ii) Shaded parts to total parts = 4 : 5
iii) Unshaded parts to total parts = 1: 5

Try These

Question 1.
Complete the following table.
TS 6th Class Maths Solutions Chapter 11 Ratio and Proportion InText Questions 5
Answer:
TS 6th Class Maths Solutions Chapter 11 Ratio and Proportion InText Questions 6

Question 2.
TS 6th Class Maths Solutions Chapter 11 Ratio and Proportion InText Questions 7
Answer:
TS 6th Class Maths Solutions Chapter 11 Ratio and Proportion InText Questions 8

Question 3.
In the following figures express the ratio of shaded parts to unshaded parts in the simplest terms.
TS 6th Class Maths Solutions Chapter 11 Ratio and Proportion InText Questions 9
Answer:
Number of shaded parts = 4
Number of unshaded parts = 12
Ratio of shaded parts to unshaded parts = 4:12
= 1:3

TS 6th Class Maths Solutions Chapter 11 Ratio and Proportion InText Questions 10
Answer:
Number of shaded parts = 3
Number of unshaded parts = 6
Ratio of shaded parts to unshaded parts = 3:6
= 1:2

TS 6th Class Maths Solutions Chapter 11 Ratio and Proportion InText Questions 11
Answer:
Number of shaded parts = 4
Number of unshaded parts = 4
Ratio of shaded parts to unshaded parts = 4 : 4
= 1:1

TS 6th Class Maths Solutions Chapter 11 Ratio and Proportion InText Questions 12
Answer:
Number of shaded parts = 6
Number of unshaded parts = 6
Ratio of shaded parts to unshaded parts = 6:6

TS 6th Class Maths Solutions Chapter 11 Ratio and Proportion InText Questions 13
Answer:
Number of shaded parts = 8
Number of unshaded parts = 12
Ratio of shaded parts to unshaded parts = 8 : 12
= 2:3

TS 6th Class Maths Solutions Chapter 11 Ratio and Proportion InText Questions 14
Answer:
Number of shaded parts = 3
Number of unshaded parts = 6
Ratio of shaded parts to unshaded parts = 3:6
= 1:2

TS 6th Class Maths Solutions Chapter 11 Ratio and Proportion InText Questions

Try These

Question 1.
Make a pattern with squared tiles using black and white tiles In the ratIo 2:5. There are many possible ways.
Answer:
TS 6th Class Maths Solutions Chapter 11 Ratio and Proportion InText Questions 15
One of the way is 2:5
2 × 5 : 5 × 5
10 : 25

Try These

Question 1.
i) In the given square rule paper 5 squares, colour 3 squares red and 2 squares green.
ii) If 10 squares are given, find how many are to be red and how maiy of them are to be green so that it becomes proportionate to the figure.
iii) If there are 15 squares then colour them accordingly.
TS 6th Class Maths Solutions Chapter 11 Ratio and Proportion InText Questions 16
Answer:
Student Activity.

TS 10th Class Maths Solutions Chapter 12 Applications of Trigonometry InText Questions

Students can practice 10th Class Maths Textbook SSC Solutions Telangana Chapter 12 Applications of Trigonometry InText Questions to get the best methods of solving problems.

TS 10th Class Maths Solutions Chapter 12 Applications of Trigonometry InText Questions

Do This

Question 1.
Draw diagram for the following situations :
i) A person is flying a kite at an angle of elevation a and the length of thread from his hand to kite is ‘l’.
ii) A person observes two banks of a river at angles of depression θ1 and θ21 < θ2) from the top of a tree of height ‘h’ which is at a side of the river. The width of the river is ‘d’. (AS5) (Page No. 297)
Solution:
i)
TS 10th Class Maths Solutions Chapter 12 Applications of Trigonometry InText Questions 1
In the figure
A is the position of the person.
B is the position of the kite.
AB is the length of thread.

ii)
TS 10th Class Maths Solutions Chapter 12 Applications of Trigonometry InText Questions 2
In the figure,
‘D’ is the position of person
CD is the height of the tree
AB is the width of the river
Angles of depression are θ1 and θ2.

TS 10th Class Maths Solutions Chapter 12 Applications of Trigonometry InText Questions

Think – Discuss

Question 1.
You are observing top of your school building at an angle of elevation a from a point which is at ‘d’ meter distance from foot of the building. Which trigonometric ratio would you like to consider to find the height of the building.
(AS3) (Page No. 297)
Solution:
Rough sketch
TS 10th Class Maths Solutions Chapter 12 Applications of Trigonometry InText Questions 3
I would like to use the “Tangent” to find the height of the building by the “Rough sketch”.

TS 10th Class Maths Solutions Chapter 12 Applications of Trigonometry InText Questions

Question 2.
A ladder of length ‘x’ meter is leaning against a wall making angle θ with the ground. Which trigonometric ratio would you like to consider to find the height of the point on the wall at which the ladder is touching ? (AS3) (Page No. 297)
Solution:
TS 10th Class Maths Solutions Chapter 12 Applications of Trigonometry InText Questions 4
I would like to use the “sin θ” to find the height of the point on the wall at which the ladder is touching.

TS 10th Class Maths Solutions Chapter 12 Applications of Trigonometry Optional Exercise

Students can practice 10th Class Maths Textbook SSC Solutions Telangana Chapter 12 Applications of Trigonometry Optional Exercise to get the best methods of solving problems.

TS 10th Class Maths Solutions Chapter 12 Applications of Trigonometry Optional Exercise

Question 1.
A 1.2 m tall girl spots a balloon moving with the wind in a horizontal line at a height of 88.2 m from the ground. The angle of elevation of the ballon from the eyes of the girl at any instant is 60°. After sometime, the angle of elevation reduces to 30°. Find the distance travelled by the balloon during the interval.
Solution:
TS 10th Class Maths Solutions Chapter 12 Applications of Trigonometry Optional Exercise 1
From the figure,
Let AD be the height of a tall girl standing on the horizontal line AB is 1.2 m.
Let FH = EB = 88.2 m be the height of balloon from the line AB.
At the eyes of the girl D, the angle of elevations are ∠FDC = 60° and ∠EDC = 30°
Now, FG = EC = 88.2 – 1.2 = 87 m.
Let the distance travelled by the balloon,
HB = y m and AH = x m.
∴ DG = x m and GC = ym
In right angled ∆FGD
tan 60° = \(\frac{\mathrm{FG}}{\mathrm{DG}}\)
⇒ \(\sqrt{3}\) = \(\frac{87}{\mathrm{x}}\)
⇒ x = \(\frac{87}{\sqrt{3}}\) ………. (1)
Again, in right angled ∆ECD,
tan 30° = \(\frac{\mathrm{EC}}{\mathrm{DC}}\)
⇒ \(\frac{1}{\sqrt{3}}\) = \(\frac{87}{\mathrm{DG}+\mathrm{GC}}\)
⇒ x + y = 87\(\sqrt{3}\) …………. (2)
∴ From equation (1), substituting
x = \(\frac{87}{\sqrt{3}}\) in equation (2), we get
TS 10th Class Maths Solutions Chapter 12 Applications of Trigonometry Optional Exercise 2
⇒ y = 29 × 2\(\sqrt{3}\) = 58 \(\sqrt{3}\) m.
Hence, the distance travelled by the balloon during the interval is 58 \(\sqrt{3}\) m.

TS 10th Class Maths Solutions Chapter 12 Applications of Trigonometry Optional Exercise

Question 2.
The angles of elevation of the top of a lighthouse from 3 boats A, B and C in a straight line of same side of the lighthouse are a, 2a, 3a respectively. If the distance between the boats A and B is x meters, find the height of lighthouse.
Solution:
From the figure,
Let PQ be the height of the lighthouse = h m
TS 10th Class Maths Solutions Chapter 12 Applications of Trigonometry Optional Exercise 3
A = First point of observation
B = Second point of observation
C = Third point of observation
Given, AB = x and BC = y (Not given in the text)
Exterior angle = Sum of the opposite interior angles
∠PBQ = ∠BQA + ∠BAQ and
∠PCQ = ∠CBQ + ∠CQB
∴ AB = x = QB
By applying the sine rule,
TS 10th Class Maths Solutions Chapter 12 Applications of Trigonometry Optional Exercise 4
⇒ h2 = \(\frac{x^2}{4 y^2}\) (3y – x)(x + y)
∴ h = \(\frac{x}{2 y} \sqrt{(3 y-x)(x+y)}\)
Height of lighthouse
= \(\frac{x}{2 y} \sqrt{(3 y-x)(x+y)}\) meters.

Question 3.
Inner part of a cupboard is in the cuboidical shape with its length, breadth and height in the ratio 1 : \(\sqrt{2}\) : 1. What is the angle made by the longest stick which can be inserted cupboard with its base inside ?
Solution:
TS 10th Class Maths Solutions Chapter 12 Applications of Trigonometry Optional Exercise 5
Inner part of a cupboard is in the cuboidical shape.
Ratio of length, breadth and height are
1 : \(\sqrt{2}\) : 1
In the figure,
Let AB be the length and BC be the height of the cupboard.
AC be the length of the stick.
‘θ’ be the angle of elevation of the stick making with base of the cupboard.
In ∆ABC,
tan θ = \(\frac{\mathrm{BC}}{\mathrm{AB}}\)
⇒ tan θ = \(\frac{1}{1}\)
⇒ tan θ = tan 45°
θ = 45°
∴ The angle of elevation is 45°.

TS 10th Class Maths Solutions Chapter 12 Applications of Trigonometry Optional Exercise

Question 4.
An iron spherical ball of volume 232848 cm3 has been melted and converted into a cone with the vertical angle of 120°. What are its height and base ?
Solution:
Given :
AC = Slant height = l
AB = Vertical height = h
BC = radius = r
Volume of iron spherical ball
V = 232848 cm3
TS 10th Class Maths Solutions Chapter 12 Applications of Trigonometry Optional Exercise 6
As per the problem,
Volume of spherical ball = Volume of cone
∴ \(\frac{1}{3}\) πr2h = 232848
In ∆ABC,
tan 60° = \(\frac{\mathrm{BC}}{\mathrm{AB}}\)
\(\sqrt{3}\) = \(\frac{\mathrm{r}}{\mathrm{h}}\) ⇒ r = \(\sqrt{3}\) h
\(\frac{1}{3}\) π(\(\sqrt{3}\) h)2 × h = 232848
\(\frac{1}{3}\) × \(\frac{22}{7}\) × 3 × h2 × h = 232848
h3 = \(\frac{232848 \times 7}{22}\)
h3 = 10584 × 7 = 74088
h3 = 423
h = 42
But r = h\(\sqrt{3}\)
∴ r = 42\(\sqrt{3}\)

TS 10th Class Maths Solutions Chapter 12 Applications of Trigonometry Optional Exercise

Question 5.
A right circular cylindrical tower, height ‘h’ and radius ‘r’, stands on the ground. Let ‘P’ be a point in the horizontal plane ground and ABC be the semi-circular edge of the top of the tower such that B is the point in it nearest to P. The angles of elevation of the points A and B are 45° and 60° respectively. Show that \(\frac{\mathrm{h}}{\mathrm{r}}=\frac{\sqrt{3}(1+\sqrt{5})}{2}\).
Solution:
As shown in the figure
TS 10th Class Maths Solutions Chapter 12 Applications of Trigonometry Optional Exercise 7
OA = BD = h (height of cylinder)
OD = r (radius of cylinder)
‘O’ is the centre.
ABC is the edge of semi-circle (in the top of cylinder).
B is nearer to the point R So B should be at the outer edge of diameter. That means just above ‘D’.
∠DPB = 60°, ∠OPA = 45° (given)
In ∆BDP, tan P = \(\frac{\mathrm{BD}}{\mathrm{DP}}\) (P = 60°, BD = h)
tan 60° = BD/DP
⇒ \(\sqrt{3}\) = \(\frac{\mathrm{h}}{\mathrm{DP}}\)
⇒ h = \(\sqrt{3}\) DP …………… (1)
In ∆OAP, tan P = \(\frac{\mathrm{OA}}{\mathrm{OP}}\) (here P = 45°, OA = h)
⇒ tan 45° = \(\frac{\mathrm{OA}}{\mathrm{OP}}\) = 1 ⇒ OA = OP
So OA = h = OP = OD + DP
So h = r + DP …………. (2)
From the equation (1) & (2)
h = \(\sqrt{3}\) DP, h = r + DP
∴ \(\sqrt{3}\) DP = r + DP
So r = \(\sqrt{3}\) DP – DP
r = DP(\(\sqrt{3}\) – 1) …………. (3)
TS 10th Class Maths Solutions Chapter 12 Applications of Trigonometry Optional Exercise 8
(But in text book, it is asked as \(\frac{\sqrt{3}(1+\sqrt{5})}{2}\) which is wrong).

TS 10th Class Maths Solutions Chapter 12 Applications of Trigonometry Ex 12.2

Students can practice 10th Class Maths Textbook SSC Solutions Telangana Chapter 12 Applications of Trigonometry Ex 12.2 to get the best methods of solving problems.

TS 10th Class Maths Solutions Chapter 12 Applications of Trigonometry Exercise 12.2

Question 1.
A TV tower stands vertically on the side of a road. From a point on the other side directly opposite to the tower, the angle of elevation of the top of tower is 60°. From another point 10 m away from this point, on the line joining this point to the foot of the tower, the angle of elevation of the top of the tower is 30°. Find the height of the tower and the width of the road. (AS4)
Solution:
In the adjacent figure,
AB denotes the height of the tower.
BC denotes the width of the road.
CD = 10 cm
∠ACB = 60° and ∠ADC = 30°
In ∆ACB, ∠B = 90°
tan 60° = \(\frac{\mathrm{AB}}{\mathrm{BC}}\)
TS 10th Class Maths Solutions Chapter 12 Applications of Trigonometry Ex 12.2 1
\(\frac{\sqrt{3}}{1}\) = \(\frac{\mathrm{AB}}{\mathrm{BC}}\) ………….. (1)
In ∆ ABD, ∠ABD = 90°
tan 30° = \(\frac{\mathrm{AB}}{\mathrm{BD}}\)
\(\frac{1}{\sqrt{3}}\) = \(\frac{\mathrm{AB}}{\mathrm{BC}+\mathrm{CD}}\)
\(\frac{1}{\sqrt{3}}\) = \(\frac{\mathrm{AB}}{\mathrm{BC}+10}\) ………….. (2)
dividing eq. (1) by eq. (2) we have
\(\frac{\mathrm{eq} \mathrm{(1)}}{\mathrm{eq} \mathrm{(2)}}=\frac{\sqrt{3} \times \sqrt{3}}{1}=\frac{\mathrm{AB}}{\mathrm{BC}} \times \frac{\mathrm{BC}+10}{\mathrm{AB}}\)
\(\frac{3}{1}\) = \(\frac{\mathrm{BC}+10}{\mathrm{BC}}\)
3BC = BC + 10
3BC – BC = 10
2BC = 10
BC = \(\frac{10}{2}\) = 5 …………… (3)
from (1),
\(\sqrt{3}\) = \(\frac{\mathrm{AB}}{\mathrm{BC}}\)
\(\frac{\sqrt{3}}{1}\) = \(\frac{\mathrm{AB}}{5}\)
⇒ AB = 5\(\sqrt{3}\)
Hence, the height of the tower = 5 \(\sqrt{3}\) m and the width of the road = 5 m.

TS 10th Class Maths Solutions Chapter 12 Applications of Trigonometry Ex 12.2

Question 2.
A 1.5 m tall boy is looking at the top of a temple which is 30 meter in height from a point of certain distance. The angle of elevation from his eye to the top of the crown of the temple increases from 30° to 60° as he walks towards the temple. Find the distance he walked towards the temple. (AS4)
Solution:
CE denotes the height of the boy CE = 1.5 m
AF denotes the height of the temple AF = 30 m
CB || EF
CEFB is a rectangle
∴ CE = 8F = 1.5 m
∴ AB = AF – BF
= 30 – 1.5
= 28.5 m
TS 10th Class Maths Solutions Chapter 12 Applications of Trigonometry Ex 12.2 13
In ∆ ABC, ∠B = 90°
∴ tan 30° = \(\frac{\mathrm{AB}}{\mathrm{BC}}\)
\(\frac{1}{\sqrt{3}}\) = \(\frac{28.5}{\mathrm{BC}}\)
⇒ BC = 28.5 × \(\sqrt{3}\) …………… (1)
In ∆ ABD, ∠B = 90°
tan 60° = \(\frac{\mathrm{AB}}{\mathrm{BD}}\)
⇒ \(\frac{\sqrt{3}}{1}\) = \(\frac{\mathrm{AB}}{\mathrm{BD}}\)
⇒ \(\frac{\sqrt{3}}{1}\) = \(\frac{28.5}{\mathrm{BD}}\)
⇒ BD × \(\sqrt{3}\) = 28.5
⇒ BD = \(\frac{28.5}{\sqrt{3}}\)
Therefore, the distance, the boy walked towards the temple is CD
CD = BC – BD
TS 10th Class Maths Solutions Chapter 12 Applications of Trigonometry Ex 12.2 3

TS 10th Class Maths Solutions Chapter 12 Applications of Trigonometry Ex 12.2

Question 3.
A statue stands on the top of a 2m tall pedestal. From a point on the ground, the angle of elevation of the top of the statue is 60° and from the same point, the angle of elevation of the top of the pedestal is 45°. Find the height of the statue. (AS4)
Solution:
In ∆ABD, ∠B = 90° and ∠DAB = 45°
∴ tan 60° = \(\frac{\mathrm{BC}}{\mathrm{AB}}\)
\(\frac{\sqrt{3}}{1}\) = \(\frac{\mathrm{BC}}{\mathrm{AB}}\)
AB × \(\sqrt{3}\) = BC
TS 10th Class Maths Solutions Chapter 12 Applications of Trigonometry Ex 12.2 4
AB = \(\frac{\mathrm{BC}}{\sqrt{3}}\) …………….. (1)
In ∆ABD, ∠B = 90° and ∠DAB = 45°
∴ tan 45° = \(\frac{\mathrm{DB}}{\mathrm{AB}}\)
\(\frac{1}{1}\) = \(\frac{2}{\mathrm{AB}}\)
⇒ AB = 2 m ………………. (2)
from (i), \(\frac{\mathrm{BC}}{\sqrt{3}}\) = \(\frac{2}{1}\)
⇒ BC = 2\(\sqrt{3}\)
= 2 × 1.732 = 3.464 m
Therefore, the height of the statue
CD = BC – BD
= 3.464 – 2 = 1.464 m

Question 4.
From the top of a building, the angle of elevation of the top of a cell tower is 60° and the angle of depression to its foot is 45°. If distance of the building from the tower is 7m, then find the height of the tower. (AS4)
Solution:
From the figure,
Let AB be the height of the tower.
CD be the height of the building.
Distance between the building from the tower is 7m
TS 10th Class Maths Solutions Chapter 12 Applications of Trigonometry Ex 12.2 5
Angles of elevation and depression are
∠BDE = 60° and ∠EDA = 45° and DE = AC = 7m and DC = AE
From the right angled ∆BDE,
We have
tan 60° = \(\frac{\mathrm{BE}}{\mathrm{DE}}\)
⇒ \(\sqrt{3}\) = \(\frac{\mathrm{BE}}{7}\)
⇒ BE = 7 \(\sqrt{3}\) ……………….. (1)
From the right angled ∆ADC, we have
tan 45° = \(\frac{\mathrm{CD}}{\mathrm{AC}}\)
1 = latex]\frac{\mathrm{CD}}{7}[/latex]
⇒ CD = 7 ………………. (2)
From the equations (1) and (2)
Height of the tower = AB = AE + BE
= 7 + 7 \(\sqrt{3}\) = 7(1 + \(\sqrt{3}\))
= 7(1 + 1.732)
= 7(2.732) = 19.124 m
Hence, the height of the tower is 19.124 m.

TS 10th Class Maths Solutions Chapter 12 Applications of Trigonometry Ex 12.2

Question 5.
A wire of length 18 m had been tied with electric pole at an angle of elevation 30° with the ground. Because it was covering a long distance, it was cut and tied at an angle of elevation 60° with the ground. How much length of the wire was cut ? (AS4)
Solution:
In the figure,
Let AB be the height of the electric pole = h m.
BC be the actual length of the wire = 18 m.
X and Y are the first and second points of observations.
TS 10th Class Maths Solutions Chapter 12 Applications of Trigonometry Ex 12.2 6
Let AX = a + b and AY = b
Angles of elevations are ∠BXA = 30° and ∠BYA = 60°
TS 10th Class Maths Solutions Chapter 12 Applications of Trigonometry Ex 12.2 7
Again from the ∆ ABY
Cos 60° = \(\frac{\mathrm{AY}}{\mathrm{BY}}\)
\(\frac{1}{2}\) = \(\frac{\mathrm{b}}{\mathrm{BY}}\)
BY = 2b
BY = 2(3\(\sqrt{3}\))
= 6\(\sqrt{3}\)
= 6 (1.732)
∴ BY = 10. 3920
∴ BY = 10.39230
The length of the cut wire = BX – BY
= 18 – 10.39230
= 7.6076 m
= 7.608 m

Question 6.
The angle of elevation of the top of a build¬ing from the foot of the tower is 30° and the angle of elevation of the top of the tower from the foot of building is 60°. If the tower is 30 m high, find the height of the building. (AS4)
Solution:
From the figure,
Let ‘BC’ be the height of the tower is 30 m.
Let ‘AD’ be the height of the building is h mC
TS 10th Class Maths Solutions Chapter 12 Applications of Trigonometry Ex 12.2 8
Angle of elevation, from the bottom of building and tower as well as
∠BAC = 60° and ∠ABD = 30°
Also, let AB = x be the distance between foot of the tower and building.
In right angled ∆ABD, we have
tan 30° = \(\frac{\mathrm{AD}}{\mathrm{AB}}\)
⇒ \(\frac{1}{\sqrt{3}}\) = \(\frac{\mathrm{h}}{\mathrm{x}}\)
⇒ h = \(\frac{\mathrm{x}}{\sqrt{3}}\) ………….. (1)
Again, in right angled ∆BAC, we have
tan 60° = \(\frac{\mathrm{BC}}{\mathrm{AB}}\)
⇒ \(\sqrt{3}\) = \(\frac{30}{\mathrm{x}}\)
x = \(\frac{30}{\sqrt{3}}\) m
Substituting x = \(\frac{30}{\sqrt{3}}\) in equation (1) we get
h = \(\frac{30}{\sqrt{3}} \times \frac{1}{\sqrt{3}}\) = \(\frac{30}{3}\) = 10 m
Hence, the height of the building is 10 m.

TS 10th Class Maths Solutions Chapter 12 Applications of Trigonometry Ex 12.2

Question 7.
Two poles of equal heights are standing opposite to each other on either side of the road, which is 120 feet wide. From a point between them on the road, the angles of elevation of the top of the poles are 60° and 30° respectively. Find the height of the poles and the distances of the point from the poles. (AS4)
Solution:
AB and CD are two poles of equal height.
Let BE = x, then ED = (120 – x)
In ∆ABE, ∠B = 90°, ∠AEB = 60°
∴ tan 60° = \(\frac{\mathrm{AB}}{\mathrm{BE}}\)
TS 10th Class Maths Solutions Chapter 12 Applications of Trigonometry Ex 12.2 9
⇒ \(\frac{\sqrt{3}}{1}\) = \(\frac{\mathrm{AB}}{\mathrm{x}}\)
⇒ AB = \(\sqrt{3}\) x ……………… (1)
In ∆CDE, ∠D = 90° and ∠CED = 30°
∴ tan 30° = \(\frac{\mathrm{CD}}{\mathrm{ED}}\)
⇒ \(\frac{1}{\sqrt{3}}\) = \(\frac{\mathrm{CD}}{120-\mathrm{x}}\)
⇒ CD × \(\sqrt{3}\) = 120 – x
⇒ CD = \(\frac{120-x}{\sqrt{3}}\) ………………… (2)
from (1) & (2), we have
\(\frac{120-x}{\sqrt{3}}\) = \(\frac{\sqrt{3} x}{1}\) (∵ AB = CD)
\(\sqrt{3}\) × \(\sqrt{3}\)x = 120 – x
3x = 120 – x
3x + x = 120
4x = 120
x = \(\frac{120}{4}\) = 30
AB = \(\sqrt{3}\) x
= \(\sqrt{3}\) × 30
= 30\(\sqrt{3}\) = 30 × 1.732
= 51.96 ft
The height of the pole = 51.96 ft.
The distance of one pole AB from the point E = 30 ft.
The distance of another pole CD from the point E = 120 – 30 = 90 ft.

TS 10th Class Maths Solutions Chapter 12 Applications of Trigonometry Ex 12.2

Question 8.
The angles of elevation of the top of a tower from two points are at a distance of 4m and 9 m, find the height of the tower from the base of the tower and in the same straight line with it are complementary. (AS4)
Solution:
From the figure,
AB be the height of a tower = h m
Let the two points on the ground are ‘C’ and
‘D’, such that they make a distance 4 m and
AC = 4 m and AD = 9 m
TS 10th Class Maths Solutions Chapter 12 Applications of Trigonometry Ex 12.2 10
Angles of elevation are ∠ACB = θ and ∠ADB = 90 – θ
In the right angled ∆ABC, we have
tan θ = \(\frac{\mathrm{AB}}{\mathrm{AC}}\)
⇒ tan θ = \(\frac{\mathrm{h}}{4}\)
Again from the right angled ∆ABD, we have
tan (90 – θ) = \(\frac{\mathrm{AB}}{\mathrm{AD}}\)
⇒ cot θ = \(\frac{\mathrm{h}}{9}\)
⇒ \(\frac{1}{\tan \theta}\) = \(\frac{\mathrm{h}}{9}\)
⇒ tan θ = \(\frac{9}{\mathrm{h}}\) …………….. (2)
From the equations (1) and (2) :
\(\frac{\mathrm{h}}{4}\) = \(\frac{9}{\mathrm{h}}\)
⇒ h2 = 36
h = \(\sqrt{36}\) = 6 m
The height of the tower is 6 m.

TS 10th Class Maths Solutions Chapter 12 Applications of Trigonometry Ex 12.2

Question 9.
The angle of elevation of jet plane from a point A on the ground is 60°. After a flight of 15 seconds, the angle of elevation changes to 30°. If the jet plane is flying at a constant height of 1500 \(\sqrt{3}\) meters, find the speed of the jet plane. (\(\sqrt{3}\) = 1.732) (AS4)
Solution:
From the figure,
Let P and Q be the two positions of the plane.
TS 10th Class Maths Solutions Chapter 12 Applications of Trigonometry Ex 12.2 11
Let A’ be the point of observation.
Let ABC be the horizontal line through A.
It is given that angle of elevation of the plane Q
from a point A’ are 60° and 30° respectively.
∴ ∠PAB = 60°, ∠QAB = 30°
Constant height of jet plane = 1500 \(\sqrt{3}\) m
In the right angled ∆ABP we have
tan 60° = \(\frac{\mathrm{BP}}{\mathrm{AB}}\)
⇒ \(\sqrt{3}\) = \(\frac{1500 \sqrt{3}}{\mathrm{AB}}\)
⇒ AC = \(\frac{1500 \sqrt{3}}{\sqrt{3}}\) = 1500 m
In the right angled ∆ACQ, we have
tan 30° = \(\frac{\mathrm{CQ}}{\mathrm{AC}}\) ⇒ \(\frac{1}{\sqrt{3}}\) = \(\frac{1500 \sqrt{3}}{\mathrm{AC}}\)
⇒ AC = 1500 × \(\sqrt{3}\) × \(\sqrt{3}\)
= 1,500 × 3 = 4,500 m
From the figure
PQ = BC = AC – AB
= 4500 – 1500 = 3000 m
Thus, the plane travels 3000 m in 15 seconds.
Hence, speed of a plane = \(\frac{3000}{15}\)
= 200m/sec.

TS 10th Class Maths Solutions Chapter 12 Applications of Trigonometry Ex 12.2

Question 10.
The angle of elevation of the top of a tower from the foot of the building is 30° and the angle of elevation of the top of the building from the foot of the tower is 60°. What is the ratio of heights of tower and building ? (AS4)
Solution:
Let the height of the tower = x m
Let the height of the building = y m
TS 10th Class Maths Solutions Chapter 12 Applications of Trigonometry Ex 12.2 12
Distance between the tower and building = d m.
Angle of elevation of the top of the tower = 30°
From the figure,
tan 30° = \(\frac{\mathrm{x}}{\mathrm{d}}\)
\(\frac{1}{\sqrt{3}}\) = \(\frac{\mathrm{x}}{\mathrm{d}}\)
d = \(\sqrt{3}\)x …………….. (1)
Also tan 60° = \(\frac{\mathrm{y}}{\mathrm{d}}\)
\(\sqrt{3}\) = \(\frac{\mathrm{y}}{\mathrm{d}}\)
d = \(\frac{y}{\sqrt{3}}\) …………….. (2)
From (1) and (2)
\(\sqrt{3}\)x = \(\frac{y}{\sqrt{3}}\)
\(\frac{x}{y}=\frac{1}{\sqrt{3} \cdot \sqrt{3}}=\frac{1}{3}\)
∴ x : y = 1 : 3
∴ The ratio of heights of tower and building = 1 : 3.

TS Inter 1st Year Maths 1A Products of Vectors Important Questions

Students must practice these TS Inter 1st Year Maths 1A Important Questions Chapter 5 Products of Vectors to help strengthen their preparations for exams.

TS Inter 1st Year Maths 1A Products of Vectors Important Questions

Question 1.
If \(\overline{\mathbf{a}}=\overline{\mathbf{i}}+2 \overline{\mathbf{j}}-3 \overline{\mathbf{k}}, \overline{\mathbf{b}}=3 \overline{\mathbf{i}}-\overline{\mathbf{j}}+2 \overline{\mathbf{k}}\) then show that \(\overline{\mathbf{a}}+\overline{\mathbf{b}}, \overline{\mathbf{a}}-\overline{\mathbf{b}}\) are perpendicular.
Solution:
TS Inter 1st Year Maths 1A Products of Vectors Important Questions 1

Question 2.
If the Vectors \(\lambda \overline{\mathbf{i}}-\overline{3} \overline{\mathbf{j}}+5 \overline{\mathrm{k}}, 2 \lambda \overline{\mathrm{i}}-\lambda \overline{\mathbf{j}}-\overline{\mathrm{k}}\) are perpendicular to each other find λ.
Solution:
TS Inter 1st Year Maths 1A Products of Vectors Important Questions 2

Question 3.
If \(\overline{\mathbf{a}}\) =6 \(\overline{\mathrm{i}}+2 \overline{\mathrm{j}}+3 \overline{\mathrm{k}}\) and \(\overline{\mathbf{b}}=2 \overline{\mathbf{i}}-9 \overline{\mathrm{j}}+6 \overline{\mathrm{k}}\) then find \(\overline{\mathrm{a}}, \overline{\mathrm{b}}\) and the angle between \(\overline{\mathbf{a}}\) and \(\overline{\mathbf{b}}\).
Solution:
TS Inter 1st Year Maths 1A Products of Vectors Important Questions 3
TS Inter 1st Year Maths 1A Products of Vectors Important Questions 4

Question 4.
Let \(\overline{\mathbf{a}}=2 \overline{\mathbf{i}}+4 \overline{\mathbf{j}}-5 \overline{\mathrm{k}}, \overline{\mathbf{b}}=\overline{\mathbf{i}}+\overline{\mathbf{j}}+\overline{\mathrm{k}}\) and \( \overline{\mathbf{c}}=\overline{\mathbf{i}}+2 \overline{\mathrm{k}}\). Find unit vector in the opposite direction of \(\overline{\mathbf{a}}+\overline{\mathbf{b}}+\overline{\mathbf{c}}\)
Solution:
TS Inter 1st Year Maths 1A Products of Vectors Important Questions 5

Question 5.
Let \(\bar{a}\) and \(\bar{b}\) be non zero, non collinear vectors. If \(|\overline{\mathbf{a}}+\bar{b}|=|\overline{\mathbf{a}}-\bar{b}|\) then find the angle between \(\bar{a}\) and \(\bar{b}\).
Solution:
TS Inter 1st Year Maths 1A Products of Vectors Important Questions 6
TS Inter 1st Year Maths 1A Products of Vectors Important Questions 7

Question 6.
If \(|\overline{\mathbf{a}}|=11,|\bar{b}|=23 \text { and }|\overline{\mathbf{a}}-\overline{\mathbf{b}}|\) = 30 then find the angle between the vectors \(\overline{\mathbf{a}}, \overline{\mathbf{b}}\), and also find \(|\overline{\mathbf{a}}+\overline{\mathbf{b}}|\)
Solution:
TS Inter 1st Year Maths 1A Products of Vectors Important Questions 8

Question 7.
If \(\overline{\mathbf{a}}=\overline{\mathbf{i}}-\overline{\mathbf{j}}-\overline{\mathbf{k}}, \overline{\mathbf{b}}=2 \overline{\mathbf{i}}-3 \overline{\mathbf{j}}+\overline{\mathbf{k}}\) then find the projection vector of \(\overline{\mathbf{b}} \text { on } \bar{a}\) and its magnitude.
Solution:
Projection vector of \(\overline{\mathbf{b}} \text { on } \bar{a}\) is
TS Inter 1st Year Maths 1A Products of Vectors Important Questions 9

Question  8.
If P, Q, R, S are points whose position vectors are \(\overline{\mathbf{i}}-\overline{\mathbf{k}},-\overline{\mathbf{i}}+\mathbf{2} \overline{\mathbf{j}}, 2 \overline{\mathbf{i}}-3 \overline{\mathbf{k}}\) and 3 \(\overline{\mathbf{i}}-2 \overline{\mathbf{j}}-\overline{\mathbf{k}}\) respectively, then find the component of \(\overline{\mathbf{R S}}\) on \(\overline{\mathbf{P Q}}\).
Solution:
Let O be the origin.
TS Inter 1st Year Maths 1A Products of Vectors Important Questions 10

Question 9.
Let \(\overline{\mathbf{a}}=2 \overline{\mathbf{i}}+3 \overline{\mathbf{j}}+\overline{\mathbf{k}}, \overline{\mathbf{b}}=4 \overline{\mathbf{i}}+\overline{\mathbf{j}}\) and \(\bar{c}=\overline{\mathbf{i}}-3 \overline{\mathbf{j}}-7 \overline{\mathbf{k}}\). Find the vector \(\overline{\mathbf{r}}\) such that \(\overline{\mathbf{r}} \cdot \overline{\mathbf{a}}=9, \overline{\mathbf{r}} \cdot \overline{\mathbf{b}}=7 \text { and } \overline{\mathbf{r}} \cdot \overline{\mathbf{c}}=6\)
Solution:
Let \(\bar{r}=x \bar{i}+y \bar{j}+z \bar{k}\)
TS Inter 1st Year Maths 1A Products of Vectors Important Questions 11
TS Inter 1st Year Maths 1A Products of Vectors Important Questions 12

Question 10.
Show that the points \(2 \bar{i}-\bar{j}+\bar{k}, \bar{i}-3 \bar{j}-5 \bar{k}\) and \( 3 \bar{i}-4 \bar{j}-4 \bar{k}\) are the vertices of a right angled triangle. Also find the other angles.
Solution:
Let O be the origin and A,B,C  be the given points, then
TS Inter 1st Year Maths 1A Products of Vectors Important Questions 13
TS Inter 1st Year Maths 1A Products of Vectors Important Questions 14

Question 11.
Prove that the smaller angle O between any two diagonals of a cube is given by \(\cos ^{-1}\left(\frac{1}{3}\right)\)
Solution:
Consider a unit cube with its vertices as shown in the figure
TS Inter 1st Year Maths 1A Products of Vectors Important Questions 15
TS Inter 1st Year Maths 1A Products of Vectors Important Questions 16
Similarly we can show the result for any other two diagonals of the cube.

Question 12.
Let \(\overline{\mathbf{a}}, \overline{\mathbf{b}}, \overline{\mathbf{c}}\) be non zero mutually orthogonal
vectors if \(\mathbf{x} \overline{\mathbf{a}}+\mathbf{y} \overline{\mathbf{b}}+\mathbf{z} \overline{\mathbf{c}}=\overline{\mathbf{0}}\) then x = y = z = 0
Solution:
TS Inter 1st Year Maths 1A Products of Vectors Important Questions 17

Question 13.
If \(4 \bar{i}+\frac{2 p}{3} \bar{j}+p \bar{k}\) is parallel to the vector \(\overline{\mathbf{i}}+2 \overline{\mathbf{j}}+3 \overline{\mathbf{k}}\). find p.
Solution:
TS Inter 1st Year Maths 1A Products of Vectors Important Questions 18
TS Inter 1st Year Maths 1A Products of Vectors Important Questions 19

Question 14.
Let \(\overline{\mathbf{a}}\) and \(\overline{\mathbf{b}} \) be vectors satisfying \(|\overline{\mathbf{a}}|=|\bar{b}|=5\) and \((\overline{\mathbf{a}}, \overline{\mathbf{b}})=45^{\circ}\)
Solution:
TS Inter 1st Year Maths 1A Products of Vectors Important Questions 20

Question 15.
Let \(\overline{\mathbf{a}}, \overline{\mathbf{b}}, \overline{\mathbf{c}}\) be mutually orthogonal vectors of equal magnitudes. Prove that the vector \(\overline{\mathbf{a}}+\overline{\mathbf{b}}+\overline{\mathbf{c}}\) is equally inclined to each of \(\overline{\mathbf{a}}, \overline{\mathbf{b}}\) and \(\overline{\mathbf{c}}\), the angle of inclination being \(\cos ^{-1}\left(\frac{1}{\sqrt{3}}\right)\)
Solution:
TS Inter 1st Year Maths 1A Products of Vectors Important Questions 21
Similarly we can show the result for any other two diagonals of the cube.

Question 16.
The vectors \(\overline{\mathrm{AB}}=3 \overline{\mathrm{i}}-2 \overline{\mathrm{j}}+2 \overline{\mathrm{k}}\) and \(\overline{\mathrm{AD}}=\overline{\mathrm{i}}-2 \overline{\mathrm{k}}\) represent the adjacent sides of a parallelogram A B C D. Find the angle between the diagonals.
Solution:
TS Inter 1st Year Maths 1A Products of Vectors Important Questions 22
TS Inter 1st Year Maths 1A Products of Vectors Important Questions 23

Question 17.
For any two vectors \(\bar{a}\) and \(\bar{b}\) show that
(i) \(|\overline{\mathbf{a}} \cdot \overline{\mathbf{b}}| \leq|\overline{\mathbf{a}}||\overline{\mathbf{b}}|\) (Cauchy – Schawartz in equality)
(ii) \(|\overline{\mathbf{a}}+\overline{\mathbf{b}}| \leq|\overline{\mathbf{a}}|+|\overline{\mathbf{b}}|\)(Triangle inequality)
Solution:
TS Inter 1st Year Maths 1A Products of Vectors Important Questions 25

Question 18.
Find the area of parallelogram for which the vectors \(\overline{\mathbf{a}}=2 \overline{\mathbf{i}}-3 \overline{\mathbf{j}} \text { and } \overline{\mathbf{b}}=3 \overline{\mathbf{i}}-\overline{\mathbf{k}}\) are adjacent sides.
Solution:
The vector area of the parallelogram
TS Inter 1st Year Maths 1A Products of Vectors Important Questions 26
TS Inter 1st Year Maths 1A Products of Vectors Important Questions 27

Question 19.
Let \(\overline{\mathbf{a}}=\overline{\mathbf{i}}+\overline{\mathbf{j}}+\overline{\mathbf{k}}, \overline{\mathbf{b}}=\mathbf{2} \overline{\mathbf{i}}-\overline{\mathbf{j}}+\mathbf{3} \overline{\mathbf{k}}, \overline{\mathbf{c}}=\overline{\mathbf{i}}-\overline{\mathbf{j}}\) and \(\bar{d}=6 \bar{i}+2 \bar{j}+3 \bar{k}\). Express \(\bar{d}\) interms of \(\overline{\mathbf{b}} \times \overline{\mathbf{c}}, \overline{\mathbf{c}} \times \overline{\mathbf{a}}\) and \(\overline{\mathbf{a}} \times \overline{\mathbf{b}}\).
Solution:
TS Inter 1st Year Maths 1A Products of Vectors Important Questions 28
TS Inter 1st Year Maths 1A Products of Vectors Important Questions 29
TS Inter 1st Year Maths 1A Products of Vectors Important Questions 30
Question 20.
Show that the angle in a semicircle is a right angle.
Solution:
Let O be the centre and AOB be the diameter of the given semicircle. Let P be any point on it. Let the position vectors of A and P be taken as \(\overline{\mathrm{a}}\) and \(\overline{\mathrm{p}}\)
TS Inter 1st Year Maths 1A Products of Vectors Important Questions 31
TS Inter 1st Year Maths 1A Products of Vectors Important Questions 32

Question 21.
For any vectors \(\overline{\mathbf{a}}, \overline{\mathbf{b}}\) and \(\overline{\mathbf{c}}\) prove that \((\overline{\mathbf{a}} \times \overline{\mathbf{b}}) \times \overline{\mathbf{c}}=(\overline{\mathbf{a}} \cdot \overline{\mathbf{c}}) \overline{\mathbf{b}}-(\overline{\mathbf{b}} \cdot \overline{\mathbf{c}}) \overline{\mathbf{a}}\)
Solution:
TS Inter 1st Year Maths 1A Products of Vectors Important Questions 33
TS Inter 1st Year Maths 1A Products of Vectors Important Questions 34

Question 22.
Find the cartesian equation of the plane passing through the point (-2,1,3) and perpendicular to the vector 3 \(\overline{\mathbf{i}}+\overline{\mathbf{j}}+\mathbf{5} \overline{\mathbf{k}}\).
Solution :
TS Inter 1st Year Maths 1A Products of Vectors Important Questions 35

Question 23.
Find the cartesian equation of the plane through the point A (2, – 1, -4) and parallel to the plane 4x – 12y – 3z – 7 = 0.
Solution:
The equation of the parallel plane is 4x- 12y-3z = p
11 passes through A (2, – 1, – 4) then
4(2)-12 (- 1)-3(-4) = p
⇒ 8 + 12 + 12 = p = p = 32
The equation of the required plane is
4x- 12y-3z = 32

Question 24.
Find the angle between the planes
2x – 3y – 6z = 5 and 6x + 2y – 9z = 4.
Solution:
Equation of the plane is 2x – 3y – 6z = 5
TS Inter 1st Year Maths 1A Products of Vectors Important Questions 36

Question 25.
Find limit vector orthogonal to the vector \(3 \bar{i}+2 \bar{j}+6 \bar{k}\) and coplanar with the vectors
\(2 \overline{\mathbf{i}}+\overline{\mathbf{j}}+\overline{\mathbf{k}} \text { and } \overline{\mathbf{i}}-\overline{\mathbf{j}}+\overline{\mathbf{k}}\)
Solution:
TS Inter 1st Year Maths 1A Products of Vectors Important Questions 37
TS Inter 1st Year Maths 1A Products of Vectors Important Questions 38

Question 26.
If \(\overline{\mathbf{a}}=2 \overline{\mathbf{i}}-3 \overline{\mathbf{j}}+5 \overline{\mathbf{k}}, \overline{\mathbf{b}}=-\overline{\mathbf{i}}+4 \overline{\mathbf{j}}+2 \overline{\mathbf{k}}\) then find \(\overline{\mathbf{a}} \times \overline{\mathbf{b}}\) and unit vector perpendicular to both \(\overline{\mathbf{a}}\) and \(\overline{\mathbf{b}}\).
Solution:
TS Inter 1st Year Maths 1A Products of Vectors Important Questions 39

Question 27.
If \(\overline{\mathbf{a}}=2 \overline{\mathrm{i}}-3 \overline{\mathrm{j}}+5 \overline{\mathrm{k}}, \overline{\mathrm{b}}=-\overline{\mathrm{i}}+4 \overline{\mathrm{j}}+2 \overline{\mathrm{k}}\) then find \((\bar{a}+\bar{b})_{\times}(\bar{a}-\bar{b})\) and unit vector perpendicular to both \(\overline{\mathbf{a}}+\overline{\mathbf{b}}\) and \(\overline{\mathbf{a}}-\overline{\mathbf{b}}\).
Solution:
TS Inter 1st Year Maths 1A Products of Vectors Important Questions 40

Question 28.
If \(\overline{\mathbf{a}}, \overline{\mathbf{b}}, \overline{\mathbf{c}}, \overline{\mathbf{d}}\) are vectors such that \(\overline{\mathbf{a}} \times \overline{\mathbf{b}}=\overline{\mathbf{c}} \times \overline{\mathbf{d}}\) and \(\overline{\mathbf{a}} \times \overline{\mathbf{c}}=\overline{\mathbf{b}} \times \overline{\mathbf{d}}\) then show that the vectors \(\overline{\mathbf{a}}-\overline{\mathbf{d}}\) and \(\overline{\mathbf{b}}-\overline{\mathbf{c}}\) are parallel.
Solution:
TS Inter 1st Year Maths 1A Products of Vectors Important Questions 41

Question 29.
Find the area of the triangle formed by the two sides \(\overline{\mathbf{a}}=\overline{\mathbf{i}}+2 \overline{\mathbf{j}}+3 \overline{\mathbf{k}}\) and \(\overline{\mathbf{b}}=3 \overline{\mathbf{i}}+5 \overline{\mathbf{j}}-\overline{\mathbf{k}}\).
Solution:
TS Inter 1st Year Maths 1A Products of Vectors Important Questions 42

Question 30.
In a \(\triangle \mathrm{ABC}\) if \(\overline{\mathrm{BC}}=\overline{\mathrm{a}}, \overline{\mathrm{CA}}=\overline{\mathrm{b}}\) and \(\overline{\mathrm{AB}}=\overline{\mathbf{c}}\) then show that \(\overline{\mathbf{a}} \times \overline{\mathbf{b}}=\overline{\mathbf{b}} \times \overline{\mathbf{c}}=\overline{\mathbf{c}} \times \overline{\mathbf{a}} \cdot\)
Solution:
TS Inter 1st Year Maths 1A Products of Vectors Important Questions 43

Question 31.
Let \(\overline{\mathbf{a}}=\mathbf{2} \overline{\mathbf{i}}-\overline{\mathbf{j}}+\overline{\mathbf{k}}\) and \(\overline{\mathbf{b}}=3 \overline{\mathbf{i}}+4 \overline{\mathbf{j}}-\overline{\mathbf{k}}\). If ‘θ’ is the angle between \(\bar{a}\) and \(\bar{b}\) then find \(\sin \theta\)
Solution:
\(\bar{a}=2 \bar{i}-\bar{j}+\bar{k}, \bar{b}=3 \bar{i}+4 \bar{j}-\bar{k}\)
TS Inter 1st Year Maths 1A Products of Vectors Important Questions 44

Question 32.
Let \(\overline{\mathbf{a}}, \overline{\mathbf{b}}, \overline{\mathbf{c}}\) be such that \(\overline{\mathbf{c}} \neq \overline{\mathbf{0}}, \overline{\mathbf{a}} \times \overline{\mathbf{b}}=\overline{\mathbf{c}}, \overline{\mathbf{b}} \times \overline{\mathbf{c}}=\overline{\mathbf{a}}\). Show that \(\overline{\mathbf{a}}, \overline{\mathbf{b}}, \overline{\mathbf{c}}\) are pair wise orthogonal vectors and \(|\overline{\mathbf{b}}|=1,|\overline{\mathbf{c}}|=|\overline{\mathbf{a}}|\).
Solution:
TS Inter 1st Year Maths 1A Products of Vectors Important Questions 49
TS Inter 1st Year Maths 1A Products of Vectors Important Questions 45

Question 33.
Let \(\overline{\mathbf{a}}=2 \overline{\mathbf{i}}+\overline{\mathbf{j}}-2 \overline{\mathbf{k}}\); \(\overline{\mathbf{b}}=\overline{\mathbf{i}}+\overline{\mathbf{j}}\). If \(\bar{c}\) is a vector such that \(\overline{\mathbf{a}} \cdot \overline{\mathbf{c}}=|\overline{\mathbf{c}}|,|\overline{\mathbf{c}}-\overline{\mathbf{a}}|=2 \sqrt{2}\) and the angle between \(\overline{\mathbf{a}} \times \overline{\mathrm{b}}\) and \(\overline{\mathrm{c}}\) is 30° then find the value of \(|(\overline{\mathbf{a}} \times \overline{\mathbf{b}}) \times \overline{\mathbf{c}}|\).
Solution:
TS Inter 1st Year Maths 1A Products of Vectors Important Questions 46

Question 34.
Let \(\overline{\mathbf{a}}, \overline{\mathbf{b}}\) be two non-collinear unit vectors if \(\bar{\alpha}=\overline{\mathbf{a}}-(\overline{\mathbf{a}} \cdot \overline{\mathbf{b}}) \overline{\mathbf{b}}\) and \(\bar{\beta}=\overline{\mathbf{a}} \times \overline{\mathbf{b}}\) then show that \(|\bar{\beta}|=|\bar{\alpha}|\).
Solution:
TS Inter 1st Year Maths 1A Products of Vectors Important Questions 50

Question 35.
A non zero vector \(\overline{\mathbf{a}}\) is parallel to the line of intersection of the plane determined by the vectors \(\overline{\mathbf{i}}, \overline{\mathbf{i}}+\overline{\mathbf{j}}\) and the plane determined by the vectors \(\overline{\mathbf{i}}-\overline{\mathbf{j}}, \overline{\mathbf{i}}+\overline{\mathbf{k}}\). Find the angle between \(\bar{a}\) and the vector \(\bar{i}-2 \bar{j}+2 \bar{k}\).
Solution:
Let l be the line of intersection of planes determined by the pairs \(\bar{i}, \bar{i}+\bar{j}\) and
TS Inter 1st Year Maths 1A Products of Vectors Important Questions 51
TS Inter 1st Year Maths 1A Products of Vectors Important Questions 52

Question 36.
Let \(\overline{\mathbf{a}}=4 \overline{\mathbf{i}}+5 \overline{\mathbf{j}}-\overline{\mathbf{k}}\), \(\overline{\mathbf{b}}=\overline{\mathbf{i}}-4 \overline{\mathbf{j}}+5 \overline{\mathbf{k}}\) and \(\overline{\mathbf{c}}=3 \overline{\mathbf{i}}+\overline{\mathbf{j}}-\overline{\mathbf{k}}\). Find the vector α which is perpendicular to both \(\bar{a}\) and \(\bar{b}\) and \(\alpha \cdot \overline{\mathbf{c}}\) = 21 \(\bar{\alpha}\) is perpendicular to both \(\bar{a}\) and \(\bar{b}\)
Solution:
TS Inter 1st Year Maths 1A Products of Vectors Important Questions 53

Question 37.
For any vector \(\overline{\mathbf{a}}\) show that
\(|\overline{\mathbf{a}} \times \overline{\mathbf{i}}|^2+|\overline{\mathbf{a}} \times \overline{\mathbf{j}}|^2+|\overline{\mathbf{a}} \times \overline{\mathbf{k}}|^2=2|\overline{\mathbf{a}}|^2\)
Solution:
TS Inter 1st Year Maths 1A Products of Vectors Important Questions 54

Question 38.
If \(\bar{a}\) is a non zero vector and \(\bar{b}\) and \(\bar{c}\) are two vectors such that \(\overline{\mathbf{a}} \times \overline{\mathbf{b}}=\overline{\mathbf{a}} \times \overline{\mathbf{c}}\) and \(\overline{\mathbf{a}} \cdot \overline{\mathbf{b}}=\overline{\mathbf{a}} \cdot \overline{\mathbf{c}}\) then prove that \(\overline{\mathbf{b}}=\overline{\mathbf{c}}\)
Solution:
TS Inter 1st Year Maths 1A Products of Vectors Important Questions 55

Question 39.
Prove that the vectors \(\overline{\mathbf{a}}=2 \overline{\mathbf{i}}-\overline{\mathbf{j}}+\overline{\mathbf{k}}, \overline{\mathbf{b}}=\overline{\mathbf{i}}-3 \overline{\mathbf{j}}-5 \overline{\mathbf{k}}\) and \(\overline{\mathbf{c}}=3 \overline{\mathbf{i}}-4 \overline{\mathbf{j}}-4 \overline{\mathbf{k}}\) are coplanar.
Solution:
\(\bar{a}=2 \bar{i}-\bar{j}+\bar{k}\)
TS Inter 1st Year Maths 1A Products of Vectors Important Questions 56

Question 40.
Find the volume of the parallelopiped whose coterminous edges are represented by the vectors \(2 \bar{i}-3 \bar{j}+\bar{k}, \bar{i}-\bar{j}+2 \bar{k}\) and \(\mathbf{2} \overline{\mathbf{i}}+\overline{\mathbf{j}}-\overline{\mathbf{k}}\)
Solution:
TS Inter 1st Year Maths 1A Products of Vectors Important Questions 57

Question 41.
Show that
\(\overline{\mathbf{i}} \times(\overline{\mathbf{a}} \times \overline{\mathbf{i}})+\overline{\mathbf{j}} \times(\overline{\mathbf{a}} \times \overline{\mathbf{j}})+\overline{\mathbf{k}} \times(\overline{\mathbf{a}} \times \overline{\mathbf{k}})=2 \overline{\mathbf{a}}\) For any vector \(\overline{\mathbf{a}}\).
Solution:
TS Inter 1st Year Maths 1A Products of Vectors Important Questions 58

Question 42.
Prove that for any three vectors \(\overline{\mathbf{a}}, \overline{\mathbf{b}}, \overline{\mathbf{c}} \left[\begin{array}{lll}
\bar{b}+\bar{c} & \bar{c}+\overline{\mathbf{a}} & \overline{\mathbf{a}}+\bar{b}
\end{array}\right]=2\left[\begin{array}{lll}
\overline{\mathbf{a}} & \bar{b} & \overline{\mathbf{c}}
\end{array}\right]\)
Solution:
TS Inter 1st Year Maths 1A Products of Vectors Important Questions 59

Question 43.
For any three vectors \(\overline{\mathbf{a}}, \overline{\mathbf{b}}, \overline{\mathbf{c}}\) prove that
\(\left[\begin{array}{lll}
\overline{\mathbf{b}} \times \overline{\mathbf{c}} & \overline{\mathbf{c}} \times \overline{\mathbf{a}} & \overline{\mathbf{a}} \times \overline{\mathbf{b}}
\end{array}\right]=\left[\begin{array}{lll}
\overline{\mathbf{a}} & \overline{\mathbf{b}} & \overline{\mathbf{c}}
\end{array}\right]^2 \cdot(\mathbf)\)
Solution:
TS Inter 1st Year Maths 1A Products of Vectors Important Questions 60

Question 44.
Let \(\bar{a}, \bar{b}\) and \(\bar{c}\) be unit vectors such that \(\bar{b}\) is not parallel to \(\overline{\mathbf{c}}\) and \(\overline{\mathbf{a}} \times(\overline{\mathbf{b}} \times \overline{\mathbf{c}})=\frac{1}{2} \overline{\mathbf{b}}\). Find the angles made by \(\bar{a}\) with each of \(\overline{\mathbf{b}}\) and \(\overline{\mathbf{c}}\).
Solution:
TS Inter 1st Year Maths 1A Products of Vectors Important Questions 61

Question 45.
For any four vectors \(\overline{\mathbf{a}}, \overline{\mathbf{b}}, \overline{\mathbf{c}}\) and \(\overline{\mathbf{d}}\).
Prove that \((\bar{b} \times \overline{\mathbf{c}}) \cdot(\overline{\mathbf{a}} \times \overline{\mathbf{d}})+(\overline{\mathbf{c}} \times \overline{\mathbf{a}}) \cdot(\overline{\mathbf{b}} \times \overline{\mathrm{d}}) +(\overline{\mathbf{a}} \times \overline{\mathbf{b}}) \cdot(\overline{\mathbf{c}} \times \overline{\mathbf{d}})=0(\mathrm)\)
Solution:
TS Inter 1st Year Maths 1A Products of Vectors Important Questions 62
= 0

Question 46.
Find the equation of the plane passing through the points \(\mathrm{A}=(2,3,-1), \mathrm{B}=(4,5, 2)\) and C=(3,6,5).
Solution:
Let \(\overline{\mathrm{OA}}=2 \overline{\mathrm{i}}+3 \overline{\mathrm{j}}-\overline{\mathrm{k}}
\overline{\mathrm{OB}}=4 \overline{\mathrm{i}}+5 \overline{\mathrm{j}}+2 \overline{\mathrm{k}}\)
\(\overline{\mathrm{OC}}=3 \overline{\mathrm{i}}+6 \overline{\mathrm{j}}+5 \overline{\mathrm{k}}\) with respect to origin \(\mathrm{O}\).
Let P be any point on the plane passing through the points A,B,C
TS Inter 1st Year Maths 1A Products of Vectors Important Questions 63

Question 47.
Find the equation of the plane passing through the point A(3,-2,-1) and parallel to the vectors \(\bar{b}=\bar{i}-2 \bar{j}+4 \bar{k}\) and \(\overline{\mathbf{c}}=3 \overline{\mathbf{i}}+2 \overline{\mathbf{j}}-5 \overline{\mathbf{k}}\).
Solution:
The equation of the plane passing through A=(3,-2,-1) and parallel to the vectors
TS Inter 1st Year Maths 1A Products of Vectors Important Questions 64
Question 48.
Find the vector equation of the plane passing through the intersection of planes \(\overline{\mathbf{r}} \cdot(\overline{\mathbf{i}}+\overline{\mathbf{j}}+\overline{\mathbf{k}})=6\) and \(\overline{\mathbf{r}} \cdot(2 \bar{i}+3 \overline{\mathbf{j}}+4 \overline{\mathbf{k}})=-5\) and the point (1,1,1).
Solution:
TS Inter 1st Year Maths 1A Products of Vectors Important Questions 66
TS Inter 1st Year Maths 1A Products of Vectors Important Questions 65

Question 49.
Find the distance of the point (2,5,-3) from the plane \(\overline{\mathbf{r}} \cdot(6 \bar{i}-3 \bar{j}+2 \bar{k})=4 \cdot\)
Solution:
Here \(\bar{a}=\bar{i}+5 \bar{j}-3 \bar{k}, \bar{n}=6 \bar{i}-3 \bar{j}+2 \bar{k}\) and d=4
Then \(\overline{\mathrm{r}} \cdot \overline{\mathbf{n}}=\overline{\mathrm{d}}\)
The distance of the point (2,5,-3) from the given plane is
TS Inter 1st Year Maths 1A Products of Vectors Important Questions 67

Question 50.
Find the angle between the line \(\frac{x+1}{2}=\frac{y}{3}=\frac{z-3}{6}\) and the plane 10 x+2 y-11 z=3
Solution:
Let φ be the angle between the given line and normal to the plane.
Concert the above equations to vector notation,
TS Inter 1st Year Maths 1A Products of Vectors Important Questions 68

Question 51.
For any four vectors \(\overline{\mathbf{a}}, \overline{\mathbf{b}}, \overline{\mathbf{c}}\) and \(\overline{\mathbf{d}}\) show that
TS Inter 1st Year Maths 1A Products of Vectors Important Questions 69
Solution:
TS Inter 1st Year Maths 1A Products of Vectors Important Questions 70

Question 52.
Find the shortest distance between the skew
TS Inter 1st Year Maths 1A Products of Vectors Important Questions 71
Solution:

TS Inter 1st Year Maths 1A Products of Vectors Important Questions 80

The first line passes through the point A(6,2,2) and parallel to the vector \(\overline{\mathrm{b}}=\overline{\mathrm{i}}-2 \overline{\mathrm{j}}+2 \overline{\mathrm{k}}\).
The second line passes through the point C(-4,0,-1) and parallel to the vector \(\overline{\mathrm{d}}=3 \overline{\mathrm{i}}-2 \overline{\mathrm{j}}-2 \overline{\mathrm{k}}\)
Shortest distance is =\(\frac{|[\overline{\overline{A C}} \bar{b} \bar{d}]|}{|\bar{b} \times \bar{d}|}\)

TS Inter 1st Year Maths 1A Products of Vectors Important Questions 72

Question 53.
i) Show that the altitudes of a triangle are concurrent.
ii) The perpendicular bisectors of the sides of a triangle are concurrent.
Solution:
(i)
TS Inter 1st Year Maths 1A Products of Vectors Important Questions 73
Consider ΔABC . O is point of intersection of altitudes.
To prove that the three altitudes are concurrent at ‘ O ‘. We have to prove that \(\overline{\mathrm{OC}}\) is perpendicular to \(\overline{\mathrm{AB}}\)
TS Inter 1st Year Maths 1A Products of Vectors Important Questions 81
∴  \(\overline{\mathrm{OC}}\) is the third altitude which passes through ‘ O ‘.
Hence the three altitudes of the triangle are concurrent.

(ii)
TS Inter 1st Year Maths 1A Products of Vectors Important Questions 82
TS Inter 1st Year Maths 1A Products of Vectors Important Questions 75
TS Inter 1st Year Maths 1A Products of Vectors Important Questions 76

Question 54.
Show that the vector area of the quadrilateral ABCD having diagonals \(\overline{\mathrm{AC}}, \overline{\mathrm{BD}}\) is \(\frac{1}{2}(\overline{\mathrm{AC}} \times \overline{\mathrm{BD}})\)
Solution:
TS Inter 1st Year Maths 1A Products of Vectors Important Questions 77
ABCD is a quadrilateral. \(\overline{\mathrm{AC}}\) and \(\overline{\mathrm{BD}}\) are diagonals of the quadrilateral. Q is the point of intersection of diagonals.
Vector area of quadrilateral ABCD = Sum of the vector area of ΔAQB, ΔBQC, ΔCQD and ΔDQA.TS Inter 1st Year Maths 1A Products of Vectors Important Questions 83

Question 55.
Let \(\overline{\mathbf{a}}, \overline{\mathbf{b}}, \overline{\mathbf{c}}\) be unit vectors such that \(\bar{b}\) is not parallel to \(\bar{c}\) and \(\bar{a} \times(\bar{b} \times \bar{c})=\frac{1}{2} \bar{b}\). Find the angle made by the vector \(\bar{a}\) with each of the vectors \(\bar{b}\) and \(\bar{c}\).
Solution:
TS Inter 1st Year Maths 1A Products of Vectors Important Questions 79