Srinivas

Solving these TS 10th Class Maths Bits with Answers Chapter 13 Probability Bits for 10th Class will help students to build their problem-solving skills.

Probability Bits for 10th Class

Question 1.
A sample space consists of 80 elementary events that are equally likely. Probability of each of them is ………….
A) 1
B) 0
C) \(\frac{1}{80}\)
D) 80
Answer:
C) \(\frac{1}{80}\)

Question 2.
If I calculate the probability of an events as – 0.5, then
A) The probability of not happening is 0.5
B) The probability of happening is 0.5
C) The event is not going to happen
D) I made a mistake
Answer:
D) I made a mistake

Question 3.
On a multiple choice test, each question has 4 possible choices. If you make a random guess, probability that you are correct is …………..
A) \(\frac{1}{4}\)
B) 1
C) 0
D) 4
Answer:
A) \(\frac{1}{4}\)

Question 4.
A bag contains 6 red marbles, 3 blue marbles and 7 green marbles. If a marble is randomly selected from the bag, the probability that it is blue ……………
A) \(\frac{1}{6}\)
B) \(\frac{1}{3}\)
C) \(\frac{3}{16}\)
D) \(\frac{13}{16}\)
Answer:
C) \(\frac{3}{16}\)

Question 5.
If an individual is selected at random, prob ability that he has a birthday in July in 2012?
A) \(\frac{30}{365}\)
B) \(\frac{31}{365}\)
C) \(\frac{30}{366}\)
D) \(\frac{31}{366}\)
Answer:
D) \(\frac{31}{366}\)

Question 6.
When a card is picked up from a deck of cards, it should be either a red or a black card because these events are :
A) Mutually exclusive
B) Equally likely
C) Complementary
D) All of these
Answer:
C) Complementary

Question 7.
Probability of getting an even or odd number in throwing a dice is …………..
A) \(\frac{1}{2}\)
B) 1
C) 0
D) \(\frac{1}{4}\)
Answer:
B) 1

Question 8.
Probability of getting 7 on a 6 faced die when it is thrown is ……………
A) 1
B) 0
C) \(\frac{1}{6}\)
D) \(\frac{1}{7}\)
Answer:
B) 0

Question 9.
Among the following probability of an event E, P(E) = …………….
A) -0.5
B) 3
C) 0.2
D) 500%
Answer:
C) 0.2

Question 10.
Two unbiased coins are tossed simultaneously. Probability of getting atmost two heads ………….
A) \(\frac{1}{4}\)
B) \(\frac{1}{2}\)
C) 1
D) \(\frac{3}{4}\)
Answer:
D) \(\frac{3}{4}\)

Question 11.
A card is pulled from a deck of 52 cards. The probability of obtaining a club is
A) \(\frac{1}{3}\)
B) \(\frac{13}{26}\)
C) \(\frac{2}{11}\)
D) \(\frac{1}{4}\)
Answer:
D) \(\frac{1}{4}\)

Question 12.
If a coin is tossed, then the probability that a head turns up is
A) \(\frac{1}{2}\)
B) \(\frac{1}{4}\)
C) \(\frac{1}{3}\)
D) \(\frac{1}{6}\)
Answer:
A) \(\frac{1}{2}\)

Question 13.
If a die is rolled then the probability of getting an even number is
A) \(\frac{1}{6}\)
B) \(\frac{1}{3}\)
C) \(\frac{1}{2}\)
D) \(\frac{2}{5}\)
Answer:
C) \(\frac{1}{2}\)

Question 14.
If two dice are thrown simutlaneously, the probability of showing the same numbers on their faces is
A) \(\frac{1}{6}\)
B) \(\frac{1}{4}\)
C) \(\frac{1}{12}\)
D) \(\frac{1}{3}\)
Answer:
A) \(\frac{1}{6}\)

Question 15.
If a card is drawn from a deck of 52 cards the probability that it is a club card is
A) \(\frac{1}{52}\)
B) \(\frac{1}{4}\)
C) \(\frac{1}{13}\)
D) \(\frac{1}{26}\)
Answer:
B) \(\frac{1}{4}\)

Question 16.
A box contains pencils and pens. The probability of picking out a pen at random is 0.65. Then the probability of not picking a pen is
A) 0.45
B) 0.55
C) 0.65
D) 0.35
Answer:
D) 0.35

Question 17.
In a simultaneous toss of two coins, prob-ability of no tails is
A) \(\frac{1}{2}\)
B) \(\frac{1}{3}\)
C) \(\frac{1}{4}\)
D) \(\frac{3}{4}\)
Answer:
D) \(\frac{3}{4}\)

Question 18.
In a simultaneous toss of two coins, the probability of atleast one head is
A) \(\frac{1}{3}\)
B) \(\frac{2}{4}\)
C) \(\frac{3}{4}\)
D) \(\frac{1}{4}\)
Answer:
C) \(\frac{3}{4}\)

Question 19.
In a single throw of two dice, the probability of getting a total of 12 is
A) \(\frac{1}{18}\)
B) \(\frac{1}{36}\)
C) \(\frac{1}{9}\)
D) \(\frac{1}{12}\)
Answer:
B) \(\frac{1}{36}\)

Question 20.
In a single throw of two dice, the probability of getting a total of 11 is
A) \(\frac{1}{9}\)
B) \(\frac{1}{18}\)
C) \(\frac{1}{12}\)
D) \(\frac{35}{36}\)
Answer:
B) \(\frac{1}{18}\)

Question 21.
In a single throw of two dice, the probability of getting a doublet is
A) \(\frac{5}{6}\)
B) \(\frac{3}{11}\)
C) \(\frac{5}{12}\)
D) \(\frac{1}{6}\)
Answer:
D) \(\frac{1}{6}\)

Question 22.
In a single throw of two dice, the probability of getting distinct numbers is
A) \(\frac{5}{6}\)
B) \(\frac{5}{12}\)
C) \(\frac{5}{36}\)
D) \(\frac{4}{36}\)
Answer:
A) \(\frac{5}{6}\)

Question 23.
In a single throw of two dice, the probability of getting even doublet is
A) \(\frac{3}{13}\)
B) \(\frac{1}{12}\)
C) \(\frac{1}{15}\)
D) \(\frac{1}{18}\)
Answer:
B) \(\frac{1}{12}\)

Question 24.
When two dice are rolled, probability of getting odd doublet is
A) \(\frac{1}{12}\)
B) \(\frac{1}{18}\)
C) \(\frac{1}{9}\)
D) \(\frac{1}{6}\)
Answer:
A) \(\frac{1}{12}\)

Question 25.
Two dice are rolled, the probability of getting 6 as the product is
A) \(\frac{1}{18}\)
B) \(\frac{1}{12}\)
C) \(\frac{1}{9}\)
D) \(\frac{1}{6}\)
Answer:
C) \(\frac{1}{9}\)

Question 26.
A page is opened at a random from a book containing 90 pages. Then the probability of a page number is a perfect square is …………….
A) \(\frac{90}{90}\)
B) \(\frac{2}{90}\)
C) \(\frac{1}{90}\)
D) None
Answer:
C) \(\frac{1}{90}\)

Question 27.
The probability of picking a red king card from a well shuffled deck of playing cards is ……………. (A.P. June. ’15)
A) \(\frac{1}{3}\)
B) \(\frac{1}{26}\)
C) \(\frac{1}{2}\)
D) 1
Answer:
B) \(\frac{1}{26}\)

Question 28.
Getting a prime or composite number is a ………………. event (A.P. Mar. ’15)
A) mutually exclusive
B) equally likely
C) 0
D) none
Answer:
A) mutually exclusive

Question 29.
P(E) = 0.65 then P(\(\overline{\mathrm{E}}\)) = ……………. (T.S. Mar. ’15)
A) 0.25
B) 1
C) 0.35
D) 0
Answer:
C) 0.35

Question 30.
If P(E) = 0.82 then P(\(\overline{\mathrm{E}}\)) = ……………..
A) 0.18
B) 0.28
C) 0.38
D) P(E) = P(\(\overline{\mathrm{E}}\))
Answer:
A) 0.18

Question 31.
Let E, \(\overline{\mathrm{E}}\)E be the complimentary events, in a random experiment, then which of the following is true ? (T.S. Mar. ’16)
A) P(E) + P(\(\overline{\mathrm{E}}\)) = 2
B) P(E) + P(\(\overline{\mathrm{E}}\)) = 3
C) P(\(\overline{\mathrm{E}}\)) + P(E) = 1
D) P(E) + P(\(\overline{\mathrm{E}}\)) = 4
Answer:
C) P(\(\overline{\mathrm{E}}\)) + P(E) = 1

Question 32.
Two fair dice are rolled and the face values are added. The probability of getting an odd number greater than 8 is ……………..
A) \(\frac{2}{9}\)
B) \(\frac{1}{6}\)
C) \(\frac{1}{4}\)
D) \(\frac{1}{9}\)
Answer:
B) \(\frac{1}{6}\)

Question 33.
A jar contains 3 mangoes and x guavas. Two fruits are pulled from the jar without replacement. An expression that represents the probability one fruit is mango and the next fruit is guava is …………..

Answer:
B)

Question 34.
Three different greeting cards and their corresponding covers are randomly strewn about on a table. If Sita puts the greeting cards into the covers at random, the probability of correctly matching of all the greeting cards and covers is …………..
A) \(\frac{5}{6}\)
B) \(\frac{2}{3}\)
C) \(\frac{1}{6}\)
D) \(\frac{1}{9}\)
Answer:
C) \(\frac{1}{6}\)

Question 35.
If two dice are rolled at a time then the probability that the two faces show different numbers is
A) \(\frac{1}{6}\)
B) \(\frac{35}{36}\)
C) \(\frac{5}{6}\)
D) \(\frac{1}{36}\)
Answer:
C) \(\frac{5}{6}\)

Question 36.
The probability of getting a number less than 5 when a die is rolled is
A) \(\frac{4}{5}\)
B) \(\frac{2}{3}\)
C) \(\frac{3}{6}\)
D) \(\frac{1}{6}\)
Answer:
B) \(\frac{2}{3}\)

Question 37.
If a ball is drawn at random from a box containing 11 red balls, 6 white balls and 9 green balls then, the probability that the ball is not green is
A) \(\frac{9}{26}\)
B) \(\frac{17}{26}\)
C) \(\frac{11}{26}\)
D) \(\frac{6}{26}\)
Answer:
B) \(\frac{17}{26}\)

Question 38.
Which of the following are equally likely events ?
A) Getting a Head or Tail in tossing a coin.
B) In a throw of a die, getting prime or composite number.
C) Drawing a number card from 1 – 50, a number divisible by 6 or 8.
D) Picking a heart or black card from a deck of playing cards.
Answer:
A) Getting a Head or Tail in tossing a coin.

Question 39.
In a single throw of two dice, the probability of getting a total of 12 is
A) \(\frac{1}{18}\)
B) \(\frac{1}{36}\)
C) \(\frac{1}{9}\)
D) \(\frac{5}{6}\)
Answer:
B) \(\frac{1}{36}\)

Question 40.
Two dice are rolled, the probability of getting 6 as the product is
A) \(\frac{1}{18}\)
B) \(\frac{1}{12}\)
C) \(\frac{1}{9}\)
D) \(\frac{1}{6}\)
Answer:
C) \(\frac{1}{9}\)

Question 41.
The ‘event’ of getting a number less than or equal to 6 is a …………….
A) base event
B) possible event
C) element
D) sure event
Answer:
D) sure event

Question 42.
When a coin is tossed, the probability of getting a head is ………….
A) \(\frac{1}{2}\)
B) 2
C) -1
D) \(\frac{3}{2}\)
Answer:
A) \(\frac{1}{2}\)

Question 43.
From a deck of cards, a card is drawn at random then, the probability of getting a black face card is …………
A) \(\frac{9}{2}\)
B) \(\frac{1}{4}\)
C) \(\frac{3}{2}\)
D) \(\frac{3}{26}\)
Answer:
D) \(\frac{3}{26}\)

Question 44.
From a bag containing 6 red balls, 5 green balls and 3 blue balls, the probability of getting a green ball at random ……………
A) \(\frac{5}{14}\)
B) \(\frac{4}{5}\)
C) \(\frac{5}{4}\)
D) None
Answer:
A) \(\frac{5}{14}\)

Question 45.
There are 50 cards numbered from 1 to 50. A card is drawn at random, then the probability that the number on the card is divisible by 8 is …………….
A) \(\frac{25}{3}\)
B) \(\frac{3}{25}\)
C) \(\frac{19}{4}\)
D) None
Answer:
B) \(\frac{3}{25}\)

Question 46.
The probability of a certain event is ………..
A) 9
B) 7
C) 0
D) 1
Answer:
D) 1

Question 47.
Probability of an event lies between ………. and ……..
A) 0, 1
B) 2, 3
C) 7, 1
D) 4, 9
Answer:
A) 0, 1

Question 48.
P(E) + P(\(\overline{\mathrm{E}}\)) = ……………….
A) 0
B) 2
C) 1
D) None
Answer:
C) 1

Question 49.
In a box, there are 28 marbles of which x are green and the rest are white. If the probability of getting a green marble is \(\frac{2}{7}\), then number of green marbles = ………….
A) 8
B) 9
C) 10
D) 13
Answer:
A) 8

Question 50.
If E is an event whose probability is \(\frac{2}{5}\), then the probability of not E is …………
A) \(\frac{1}{2}\)
B) \(\frac{5}{3}\)
C) \(\frac{3}{5}\)
D) \(\frac{1}{3}\)
Answer:
C) \(\frac{3}{5}\)

Question 51.
If two dice are rolled simultaneously then the ‘sum’ with greatest possibility to happen is …………..
A) 71
B) 7
C) 3
D) None
Answer:
B) 7

Question 52.
The probability of raining in a day is …………..
A) \(\frac{-1}{2}\)
B) \(\frac{1}{2}\)
C) \(\frac{1}{4}\)
D) None
Answer:
B) \(\frac{1}{2}\)

Question 53.
If one side is chosen at random from the sides of a right triangle, then the probability that it is hypotenuse is ……………
A) 2
B) \(\frac{1}{2}\)
C) 3
D) \(\frac{1}{3}\)
Answer:
D) \(\frac{1}{3}\)

Question 54.
When a dice is thrown, the probability of getting neither a prime nor composite number is ……………
A) \(\frac{1}{3}\)
B) \(\frac{1}{2}\)
C) \(\frac{1}{6}\)
D) None
Answer:
C) \(\frac{1}{6}\)

Question 55.
Getting a Tail or Head ………………..
A) equally likely
B) unlikely
C) exclusive
D) None
Answer:
B) unlikely

Question 56.
Getting a prime (or) composite
A) mutually exclusive
B) likely !
C) 0
D) None
Answer:
D) None

Question 57.
Getting a red card (or) black card is ……………..
A) mutually exclusive
B) more likely
C) less likely
D) None
Answer:
A) mutually exclusive

Question 58.
P (Sure event) = ……………….
A) 1
B) 0
C) -1
D) 2
Answer:
A) 1

Question 59.
P (Impossible event) = ……………..
A) 4
B) 3
C) -1
D) 0
Answer:
D) 0

Question 60.
The probability of a face card from red cards is ……………….
A) \(\frac{3}{13}\)
B) \(\frac{13}{3}\)
C) \(\frac{2}{17}\)
D) None
Answer:
A) \(\frac{3}{13}\)

Question 61.
The probability of drawing a black king from the deck is ………………
A) \(\frac{1}{14}\)
B) \(\frac{1}{3}\)
C) \(\frac{1}{2}\)
D) \(\frac{1}{26}\)
Answer:
D) \(\frac{1}{26}\)

Question 62.
The probability of drawing a black card front he black cards is ………………
A) 3
B) 2
C) 0
D) 1
Answer:
D) 1

Question 63.
The probability of getting two tails when two coins are tossed is ……………..
A) \(\frac{1}{4}\)
B) \(\frac{1}{2}\)
C) \(\frac{2}{3}\)
D) None
Answer:
A) \(\frac{1}{4}\)

Question 64.
There are …………. cards in a pack of playing cards.
A) 19
B) 16
C) 52
D) 50
Answer:
C) 52

Question 65.
P(E) = 0.05 then P(\(\overline{\mathrm{E}}\)) = ……………
A) 1.35
B) 0.95
C) 9.5
D) 1.5
Answer:
B) 0.95

Question 66.
P(G) = \(\frac{4}{17}\), P(\(\overline{\mathrm{G}}\)) = …………..
A) \(\frac{13}{17}\)
B) \(\frac{3}{17}\)
C) \(\frac{7}{17}\)
D) \(\frac{1}{17}\)
Answer:
A) \(\frac{13}{17}\)

Question 67.
P(N) + P(\(\overline{\mathrm{N}}\)) = …………….
A) 0
B) 1
C) 3
D) 7
Answer:
B) 1

Question 68.
A baby is born the probability that it is a boy (or) girl is …………….
A) 1
B) \(\frac{-1}{2}\)
C) \(\frac{1}{3}\)
D) \(\frac{1}{2}\)
Answer:
D) \(\frac{1}{2}\)

Question 69.
P(E) + P (not E) = ………………
A) 1
B) 2
C) 3
D) None
Answer:
A) 1

Question 70.
Identify true statement. ( )
A) 0 < P(E) < 1
B) 0 < P(E) < 2
C) p < P(E)
D) None
Answer:
A) 0 < P(E) < 1

Question 71.
There are ……………… face cards.
A) 1
B) 2
C) 4
D) None
Answer:
D) None

Question 72.
Probability can never be ……………
A) 0
B) 1
C) 0.5
D) -2
Answer:
D) -2

Question 73.
A dice is tossed once then the probability of getting an even number or a multiple of 3 is
A) \(\frac{1}{2}\)
B) \(\frac{2}{3}\)
C) \(\frac{1}{4}\)
D) None
Answer:
B) \(\frac{2}{3}\)

Question 74.
The probability that a leap year has 53 Sundays is ……………….
A) \(\frac{2}{7}\)
B) \(\frac{3}{7}\)
C) \(\frac{1}{7}\)
D) \(\frac{21}{17}\)
Answer:
A) \(\frac{2}{7}\)

Question 75.
Two dice are thrown once together What is the probability of getting a doublet ?
A) \(\frac{1}{4}\)
B) \(\frac{1}{2}\)
C) \(\frac{1}{6}\)
D) None
Answer:
C) \(\frac{1}{6}\)

Question 76.
P(E) – 1 + P(\(\overline{\mathrm{E}}\)) = ………………
A) -2
B) 0
C) 9
D) 2
Answer:
B) 0

Question 77.
P(E) = 0.455 then P(\(\overline{\mathrm{E}}\)) = ……………….
A) 0.545
B) 0.145
C) 0.345
D) None
Answer:
A) 0.545

Question 78.
P(A¹) = …………………
A) Φ
B) A
C) 1 – P(A)
D) None
Answer:
C) 1 – P(A)

Question 79.
Karishma and Reshma are playing chess. The probability of winning Karishma is 0.59. Then probability of Reshma winnig the match is …………… (A.P. Mar. ’15)
A) 1
B) 0.46
C) 0.5
D) 0.41
Answer:
D) 0.41

Question 80.
Vineeta said that probability of impossible events is 1. Dhanalakshmi said that probability of sure event is ‘O’ and Sireesha said that probability of any event lies in between 0 and 1. In the above with whom will you agree? (A.P. Mar. ’15)
A) Vineetha
B) Dhanalakshmi
C) Sireesha
D) All the three
Answer:
C) Sireesha

Question 81.
From the figure the probability of getting blue ball is ……………. (A.P. Mar. ’15, ’16)

A) \(\frac{3}{5}\)
B) \(\frac{3}{3}\)
C) \(\frac{5}{5}\)
D) \(\frac{5}{3}\)
Answer:
A) \(\frac{3}{5}\)

These TS 10th Class Maths Chapter Wise Important Questions Chapter 11 Trigonometry given here will help you to solve different types of questions.

TS 10th Class Maths Important Questions Chapter 11 Trigonometry

Previous Exams Questions

Question 1.
If sin A = cos A then find the value of A. (A.P. Mar. ’15)
Solution:
sin A = cos A (given)
then A + A = 90 ⇒ 2A = 90
⇒ A = \(\frac{90}{2}\) = 45°
∴ A = 45°

Question 2.
If 4 sin² θ – 1 = 0 then find ‘θ’ (θ < 90) also, find the value of θ and the value of cos²θ + tan²θ (AP. Mar. ’15)
Solution:
Given, 4 Sin²θ – 1 = 0
4 Sin²θ = 1
Sin²θ = \(\frac{1}{4}\)
Sinθ = ± \(\sqrt{\frac{1}{4}}\) = ±\(\frac{1}{2}\)
Given θ is less than 90°
∴ Sin θ = \(\frac{1}{2}\)
sinθ = sin 30°
∴ θ = 30°
Cosθ = Cos 30° = \(\frac{\sqrt{3}}{2}\)
Tan θ = Tan 30° = \(\frac{1}{\sqrt{3}}\)
Cos² θ + Tan² θ = Cos² 30° + Tan² 30°
= \(\left(\frac{\sqrt{3}}{2}\right)^2\) + \(\left(\frac{1}{\sqrt{3}}\right)^2\)
= \(\frac{3}{4}\) + \(\frac{1}{3}\) = \(\frac{9+4}{12}\) = \(\frac{13}{12}\)

Question 3.
Show that tan²θ – \(\frac{1}{\cos ^2 \theta}\) = 1 (T.S. Mar. ’15)
Solution:
Method – I : Since \(\frac{1}{\cos ^2 \theta}\) = sec²θ

Question 4.
Find the value of (T.S. Mar. ’15)

Solution:

Question 5.
If tan θ = \(\sqrt{3}\) (θ is acute angle) then find the value of 1 + cos θ. (T.S. Mar. ’16)
Solution:
tan θ = \(\sqrt{3}\) = tan 60 (∵ θ is acute)
⇒ θ = 60
⇒ 1 + cos θ = 1 + cos 60 = 1 + \(\frac{1}{2}\) = \(\frac{3}{2}\)
∴ 1 + cos θ = \(\frac{3}{2}\)

Question 6.
Show that \(\sqrt{\frac{1-\sin \theta}{1+\sin \theta}}\) = sec θ – tan θ. (T.S. Mar. ’16)
Solution:
\(\sqrt{\frac{1-\sin \theta}{1+\sin \theta}}\) = \(\sqrt{\frac{(1-\sin \theta)(1-\sin \theta)}{(1+\sin \theta)(1-\sin \theta)}}\)

Hence proved

Question 7.
If tan (A + B) = 1, and cos (A – B) = \(\frac{\sqrt{3}}{2}\) 0° < A + B < 90, A > B then find values of A and B. (T.S. Mar. ’16)
Solution:
tan (A + B) = 1 = tan 45°
∴ A + B = 45° ……………. (1)
cos (A – B) = \(\frac{\sqrt{3}}{2}\) = cos 30°
⇒ A – B = 30° ……………… (2)
Solving the equation (1) and (2) we get

then A + B = 45
37.5 + B = 45 ⇒ B = 45 – 37.5 = 7.5
So, A = 37.5°, B = 7.5°

Question 8.
Find the value of tan² 60 + 4 cos² 45 + 3 sec² 30 + 5 cos² 90 = cosec 30 + sec 60 – cot² 30 (T.S. Mar. ’16)
Solution:
Put the following values in the given problem
tan 60° = \(\sqrt{3}\) , cos 45° = \(\sqrt{2}\), sec 30° = \(\frac{2}{\sqrt{3}}\)
cos 90° = 0, cosec 30° = 2, sec 60° = 2, cot 30° = \(\sqrt{3}\)
We get
tan² 60 + 4 cos² 45 + 3 sec² 30 + 5 cos² 90 = cosec 30 + sec 60 – cot² 30

Additional Questions

Question 1.
In a right angled triangle ABC, with right angle at B in which a = 5 units, b = 13 units and ∠BCA = θ, then find sin θ and tan θ.
Solution:
Given a = 5 units = BC
b = 13 units = CA or AC
∠BCA = θ
By pythagoras theorem
AC² = AB² + BC²
⇒ 13² = AB² + 5²
⇒ AB² = 13² – 5² = 169 – 25 = 144
⇒ AB = \(\sqrt{144}\) = 12 units
Now from the figure
Sin θ = \(\frac{\mathrm{AB}}{\mathrm{AC}}\) = \(\frac{12}{13}\)
and Tan θ = \(\frac{\mathrm{AB}}{\mathrm{AC}}\) = \(\frac{12}{5}\)

Question 2.
If Cos c = \(\frac{3}{5}\), then find Sin c and Tan c
Solution:
We have cos c = \(\frac{3}{5}\) = \(\frac{\mathrm{BC}}{\mathrm{AC}}\)

⇒ AC = 5, BC = 3
By Pythagoras theorem
AC² = AB² + BC²
⇒ 5² = AB² + 3²
⇒ 25 = AB² + 9 = 16
⇒ AB = \(\sqrt{16}\) = 4
From Sin c = \(\frac{\mathrm{AB}}{\mathrm{AC}}\) = \(\frac{4}{5}\), Tan c = \(\frac{\mathrm{AB}}{\mathrm{AC}}\) = \(\frac{4}{3}\)

Question 3.
If 12 Tan A = 9, then find Sin A and Cos A.
Solution:
Given 12 Tan A = 9

⇒ Tan A = \(\frac{9}{12}\) = \(\frac{\mathrm{BC}}{\mathrm{AB}}\)
⇒ BC = 9, AB = 12
By Pythagoras theorem
AC² = AB² + BC²
= 12² + 9²
= 144 + 81
AC² = 225
AC = \(\sqrt{225}\) = 15
Now, sin A = \(\frac{\mathrm{BC}}{\mathrm{AC}}\) = \(\frac{9}{5}\) and cos A = \(\frac{\mathrm{AB}}{\mathrm{AC}}\) = \(\frac{12}{15}\)
∴ sin A = \(\frac{9}{15}\) and cos A = \(\frac{12}{15}\)

Question 4.
If 5Cot A = 12, find Cos A and Cosec A.
Solution:
Given Cot A = \(\frac{12}{5}\) = \(\frac{\mathrm{AB}}{\mathrm{BC}}\)
⇒ AB = 12, BC = 5
By Pythagoras theorem
AC² = AB² + BC²
= 12² + 5²
= 144 + 25
= AC² = 169
AC = \(\sqrt{169}\) = 13
From the figuref ∆ABC,
Cos A = \(\frac{\mathrm{AB}}{\mathrm{AC}}\) = \(\frac{12}{13}\), Cosec A = \(\frac{\mathrm{AC}}{\mathrm{BC}}\) = \(\frac{13}{5}\)
∴ Cos A = \(\frac{12}{13}\) and Cosec A = \(\frac{13}{5}\)

Question 5.
Evaluate the following.
i) Sin 60° + Cos 60°
Solution:
Sin 60° + Cos 60° = \(\frac{\sqrt{3}}{2}\) + \(\frac{1}{2}\) = \(\frac{\sqrt{3}+1}{2}\)

ii) \(\frac{{Sin} 45^{\prime \prime}}{{Sin} 30^{\prime \prime}+{Cos} 60}\)
Solution:

iii) Tan² 30° + Cot² 45° – Cos 60
Solution:
Tan² 30° + Cot² 45° – Cos 60
= \(\left(\frac{1}{\sqrt{3}}\right)^2\) + (1)² – \(\frac{1}{2}\)
= \(\frac{1}{3}\) + \(\frac{1}{1}\) – \(\frac{1}{2}\)
= \(\frac{2+6-3}{6}\)
= \(\frac{5}{6}\)

iv) 2 Tan² 45° + Sin² 60° – Cos² 30°
Sol:
2 Tan² 45° + Sin² 60° – Cos² 30°
= 2(1)² + \(\left(\frac{\sqrt{3}}{2}\right)^2\) – \(\left(\frac{\sqrt{3}}{2}\right)^2\)
= 2 × 1 + \(\frac{3}{4}\) – \(\frac{3}{4}\) = 2

v) Cot² 30° + 4 Sin² 45° + 3 Cosec 60°
Solution:
Cot² 30° + 4 Sin² 45° + 3 Cosec 60°
= (\(\sqrt{3}\))² + 4\(\left(\frac{1}{\sqrt{2}}\right)^2\) + 3\(\left(\frac{2}{\sqrt{3}}\right)^2\)
= 3 + 4 × \(\frac{1}{2}\) + 3 × \(\frac{4}{3}\)
= 3 + 2 + 4
= 9

vi) \(\sqrt{2}\) Sin 45° + Cos 90° + Sin 90°
Solution:
\(\sqrt{2}\) .Sin 45° + Cos 90° + Sin 90°
= \(\sqrt{2}\) . \(\frac{1}{\sqrt{2}}\) + 0 + 1
= 1 + 1
= 2

Question 6.
Evaluate Cos 60°, Cos 30° – Sin 60° Sin 30°
What is the value of Cos (60° + 30°) ? What can you conclude ?
Solution:
Take Cos 60° Cos 30° – Sin 60°. Sin 30°
= \(\frac{1}{2}\) \(\frac{\sqrt{3}}{2}\) – \(\frac{\sqrt{3}}{2}\) . \(\frac{1}{2}\)
= \(\frac{\sqrt{3}}{4}\) – \(\frac{\sqrt{3}}{4}\)
= 0 …………….. (1)
Now take Cos(60° + 30°) = Cos (90°) = 0 ………….. (2)
From equations (1) and (2), I conclude that
Cos(60° + 30°) = Cos 60° . Cos 30° – Sin 60°. Sin 30°
i.e.,Cos(A + B) = Cos A . Cos B – Sin A . Sin B.

Question 7.
Is it right to say Sin (60° – 30°) = Sin 60°. Cos 30° – Cos 60°. Sin 30° ?
Solution:
LHS = Sin(60° – 30°)
= Sin 30° = \(\frac{1}{2}\)
RHS = Sin 60°. Cos 30° – Cos 60°. Sin 30°
= \(\frac{\sqrt{3}}{2}\) . \(\frac{\sqrt{3}}{2}\) – \(\frac{1}{2}\) . \(\frac{1}{2}\)
= \(\frac{3}{4}\) – \(\frac{1}{4}\) = \(\frac{-1}{4}\)
= \(\frac{2}{4}\)
= \(\frac{1}{2}\)
Yes, it is right to say
Sin(60° – 30°) = Sin 60°. Cos 30° – Cos 60°. Sin 30°
i.e., Sin (A – B) = Sin A Cos B – Cos A. Sin B.

Question 8.
Is it right to say that Cos (A + B) = Cos A + Cos B ?
Solution:
Take A = 60°, B = 30°
Then Cos(A + B) = Cos (60° + 30°)
= Cos 90° = 0
Cos A + Cos B = Cos 60° + Cos 30°
= \(\frac{1}{2}\) + \(\frac{\sqrt{3}}{2}\)
= \(\frac{1+\sqrt{3}}{2}\)
∴ Cos (A + B) ≠ Cos A + Cos B
It is not right to say that Cos (A + B) ≠ Cos A + Cos B

Question 9.
i) \(\frac{{Sin} 66^{\prime \prime}}{{Cos} 24 “}\)
ii) \(\frac{{Sin} 18^{\prime \prime}}{{Cos} 72^{\prime \prime}}\)
iii) \(\frac{{Tan} 80^{\prime \prime}}{{Cot} 10^{\prime \prime}}\)
iv) \(\frac{{Sec} 69^{\prime \prime}}{{Cosec} 21 “}\)
v) \(\frac{{Tan} 54 “}{{Cot} 36^{\prime \prime}}\)
Solution:

Question 10.
Find the value of
i) Sin 75° – Cos 15°
ii) Sec 23° – Cosec 67°
iii) Sec 70° – Sec 20°
iv) Tan 68° – Tan 22°
Solution:
i) Given Sin 75° – Cos 15°
= Sin 75° – Cos(90 – 75°)
= Sin 75° – Sin 75°
= 0

ii) Given Sec 23° – Cosec 67°
= Sec 23° – Cosec (90 – 23°)
= Sec 23° – Sec 23°
= 0

iii) Sec 70° Sec 20°
= Sin 70° . \(\frac{1}{\cos 20 “}\) = \(\frac{{Sin} 70^{\prime \prime}}{{Cos}\left(90^{\prime \prime}-70^{\prime \prime}\right)}\)
= \(\frac{{Sin} 70 “}{{Cos}(90 “-70 “)}\) = \(\frac{{Sin} 70^{\prime \prime}}{{Sin} 70^{\prime \prime}}\) = 1

iv) Given Tan 68°. Tan 22°
Tan 68° . Tan(90° – 68°)
Tan 68° . Cot 68°
= \(\frac{{Tan} 68^{\prime \prime}}{{Tan} 68^{\prime \prime}}\) = 1

Question 11.
If Cot 2A = Tan (A – 18), when 2A is an acute angle, find the value of A.
Solution:
Given that Cot 2A = Tan (A – 18°)
⇒ Tan (90° – 2A) = Tan (A – 18°) [∵ Cot θ = Tan (90 – θ)]
⇒ 90 – 2A = A – 18°
⇒ 90 + 18° = A + 2A
⇒ 3A = 108°
⇒ A = \(\frac{108 “}{3}\) = 36°
Hence, the value of A is 36°.

Question 12.
If Cos 4A = Sin(A – 20), when 4A is an acute angle, find the value of A.
Solution:
Given that Cos 4A = Sin(A – 20°)
⇒ Sin(90° – 4A) = Sin(A – 20°)
[∴ Cos 0 = Sin(90 – 0)]
⇒ 90 – 4A = A – 20°
⇒ 90 + 20 = A + 4A
⇒ 110° = 5A
⇒ A = \(\frac{110 “}{15}\) = 22°
Hence, the value of A is 22°.

Question 13.
If Sin θ + Cosec θ = 2, find the value of Sin² θ + Cosec² θ
Solution:
Given Sin θ + Cosec θ = 2
Squaring on both sides
(Sin θ + Cosec)² = 2²
⇒ Sin² θ + Cosec² θ + 2 . Sin θ . Cosec θ = 4
⇒ Sin² θ + Cosec² θ + 2 . Sin θ . \(\frac{1}{{Sin} \theta}\) = 4
⇒ Sin² θ + Cosec² θ + 2 = 4
⇒ Sin² θ + Cosec² θ + 4 – 2
⇒ Sin² θ + Cosec² θ = 2

Question 14.
Show that \(\frac{{Tan} A+{Cot} B}{{Tan} B+{Cot} A}\) = Tan A . Cot B.
Solution:

= Cot B . Tan A
= Tan A . Cot B = RHS
LHS = RHS

Question 15.
Show that (Sin θ + Cos θ)² – (Sin θ – Cos θ)² = 4 Sin θ Cos θ
Solution:
LHS = (Sin θ + Cos θ)² – (Sin θ – Cos θ)²
= (Sin² θ + Cos² θ + 2 Sin θ . Cos θ) – (Sin² θ + Cos² θ – 2 Sin θ . Cos θ)
= (2 Sin θ + Cos θ) – (-2 Sin θ . Cos θ)
= 2 Sin θ + Cos θ + 2 Sin θ . Cos θ
= 4 Sin θ . Cos θ
= RHS
∴ LHS = RHS

Question 16.
Find the value of \(\frac{{Tan}^2 60^{\prime \prime}+4 {Sin}^2 45^{\prime \prime}+3 {Sec}^2 30^{\prime \prime}+10 {Cos}^2 90^{\prime \prime}}{{Cosec}^2 60^{\prime \prime}+{Cos} 60^{\prime \prime}-{Cot}^2 60^{\prime \prime}}\)
Solution:
Put the following values in the given problem.

Question 17.
Show that \(\frac{1}{{Sin} \theta}\) – Sin θ = Cot θ. Cos θ
Solution:

Question 18.
Show that \(\frac{{Tan}^2 q}{1+{Sec} q}=\frac{1-{Cos} q}{{Cos} q}\)
Solution:

∴ LHS = RHS

Question 19.
Evaluate : log₄ (1 + tan² 45°)²
Solution:
∴ log₄ (1 + 1)²
= log₄ = 1

Question 20.
Is it true to say that cos (60°+30°) = cos 60° cos 30° + sin 60° sin 30°
Solution:
Here L.H.S = cos (60° + 30°) = cos 90° = 0
R.H.S – cos 60° cos 30° + sin 60° sin 30°
= \(\frac{1}{2}\) × \(\frac{\sqrt{3}}{2}\) + \(\frac{\sqrt{3}}{2}\) × \(\frac{1}{2}\)
= \(\frac{\sqrt{3}}{4}\) + \(\frac{\sqrt{3}}{4}\)
= \(\frac{\sqrt{3}}{4}\)
= \(\frac{\sqrt{3}}{2}\)
Here L.H.S ≠ R.H.S, so it’s not justify.

Question 21.
In ∆ABC, ∠C = 90° If BC + CA = 17 cm; BC – CA = 7 cm. Find
(i) sin A
(ii) sin B
Solution:
Here, given ∠C = 90°; BC + CA = 7 cm;
BC – CA = 7 cm

BC = \(\frac{24}{2}\) = 12 cm; figure script pno.4
BC = 12 cm
We apply BC = 12 cm. In BC + CA = 17 cm
Then 12 + CA = 17
CA = 17 – 12 = 5 cm
We know from ∆ABC, AB² = BC² + CA²
AB = \(\sqrt{\mathrm{BC}^2+\mathrm{CA}^2}\)
AB = \(\sqrt{12^2+5^2}\) = \(\sqrt{144+5}\)
AB = \(\sqrt{169}\) = 13
We know = sin A = \(\frac{\mathrm{BC}}{\mathrm{AB}}\) = \(\frac{12}{13}\)
sin B = \(\frac{\mathrm{AC}}{\mathrm{AB}}\) = \(\frac{5}{13}\)

Question 22.
Find the value of cos² 1° + cos² 2° + cos² 3° + ……………. + cos² 90°
Solution:
We know cos² (90° – 89°) + cos² (90° – 88°) + …………… + cos² 89° + cos² 90°
Here total 90° terms is there we know sin²θ + cos²θ = 1
= (sin² 89° + cos² 89) + (sin² 89 + cos² 88) + ………….. 44 terms.
= 44(1) + \(\frac{1}{\sqrt{2}}\) + 1
= 45 + \(\frac{1}{\sqrt{2}}\)

Question 23.
If cosec θ + cot θ = k then prove that cos θ = \(\frac{\mathrm{k}^2-1}{\mathrm{k}^2+1}\).
Solution:
Given cosec θ + cot θ = k ………….. (1)
We know cosec²θ – cot²θ = 1
∴ (cosec θ + cot θ) (cosec θ – cot θ) = 1 k(cosec θ – cot θ) = 1
k(cosec θ – cot θ) = 1
cosec θ – cot θ = …………….. (2)

These TS 10th Class Maths Chapter Wise Important Questions Chapter 13 Probability given here will help you to solve different types of questions.

TS 10th Class Maths Important Questions Chapter 13 Probability

Previous Exams Questions

Question 1.
5 red and 8 white balls are present in a bag. If a ball is taken randomly from the bag then find the probability of it to be
i) white ball
ii) not to be white ball (A.P. Mar. ’16)
Solution:
Total number of balls present in bag
= 5 (red) + 8 (white) = 13
Probability for taking out a white ball
P(E) = \(\frac{\text { No. of favourable outcomes }}{\text { Total No. of outcomes }}\)
= \(\frac{8}{13}\)
Probability for not to be a white ball
P(\(\overline{\mathrm{E}}\)) = \(\frac{8}{13}\)
we know P(E) + P(\(\overline{\mathrm{E}}\)) = 1
⇒ P(\(\overline{\mathrm{E}}\))
= 1 – P(E) = 1 – \(\frac{8}{13}\)
= \(\frac{5}{13}\)

Question 2.
When die is rolled once unbiased what is the probability of getting a multiple of 3 out of possible out comes ? (T.S. Mar. ’15)
Solution:
P(E) = \(\frac{\text { favourable outcomes }}{\text { Total outcomes }}\) = \(\frac{2}{6}\) = \(\frac{1}{3}\)

Question 3.
There are 12 red, 18 blue and 6 white balls in a box. When balls is drawn at random from the box, what is the probability of not getting a red ball ? (T.S. Mar. ’15)
Solution:
Total Number of balls = 12 + 8 + 6
= 36
Number of red balls = 12
probability of getting red ball
P(\(\overline{\mathrm{R}}\)) = \(\frac{\text { favourable outcomes }}{\text { Total outcomes }}\)
= \(\frac{12}{36}\) = \(\frac{1}{3}\)
∴ Probability of not getting red ball
P(R) = 1 – \(\frac{1}{3}\) = \(\frac{2}{3}\)
Total Number of balls = 12 + 18 + 6 = 36
Exclude, the red balls, the number of remaining balls = 18 + 6 = 24
∴ Probability of not getting a
Red ball = \(\frac{24}{36}\) = \(\frac{2}{3}\)

Question 4.
There are 100 flash cards labelled from 1 to 100 in a bag. When a card is drawn from the bag at random, what is the probability of getting …….
(i) a card with prime number from possible outcomes ?
(ii) a card without prime number from possible outcomes ? (T.S. Mar. ’15)
Solution:
Number of prime numbers between 1 and 100 = 25
Probability of getting a card with prime numbers = \(\frac{25}{100}\) = \(\frac{1}{4}\) = 0.25
Probability of getting a card without prime number = \(\frac{75}{100}\) = 0.75
1 – 0.25 = 0.75

Question 5.
Find the probability of setting a sum of the numbers on them is 7, when two dice are rolled at a time. (T.S. Mar. ’16)
Solution:
When two dice are rolled at a time the total outcomes are = 6² = 36.
Number of outcomes such that their sum of numbers on face is 7 = 6
∴ Probability of getting sum of numbers on faces to be
7 = \(\frac{6}{36}\) = \(\frac{1}{6}\)

Question 6.
A bag contains 5 red balls and some blue balls. If the probability of drawing a blue ball is double that of red ball, find the number of blue balls in the bag. (T.S. Mar. ’16)
Solution:
Number of red balls present in a bag = 5
Let the No. of blue balls = x (say)
Then the total No. of balls = 5 + x
From those (5 + x) balls in the bag the number of favourable outcomes to take a red ball randomly = 5
So, the probability of taking a red ball = \(\frac{5}{5+x}\)
Now
The number of favourable outcomes to take a blue ball randomly = x
So, the probability of taking a blue ball = \(\frac{x}{5+x}\)
From the given problem
Probability of blue bell = (Probability of red ball) (2)
\(\frac{x}{5+x}\) = \(\left[\frac{5}{5+x}\right]\) 2
\(\frac{x}{5+x}\) = \(\frac{10}{5+x}\) ⇒ x = 10
∴ No. of blue balls in the bag = 10.

Additional Questions

Question 1.
Kishore buys a fruit from a shop. The shopkeeper have one box. The box contain 18 mangoes, 32 apples so shopkeeper takes out one fruit at random what is the probability that the mango taken out from box.
Solution:
Given
Number of mangoes in the box = 18
Number of apples in the box = 32
Total number of fruits in the box = 18 + 32
So total number of outcomes = 50
Let E be the even that the mango taken out of the box = 18
We know that,
P(E) = \(\frac{\text { No. of outcomes favourable to E}}{\text { Total No. of all possible outcomes }}\)
= \(\frac{18}{50}\) = \(\frac{9}{25}\)

Question 2.
A room contains 30 green chairs and some white chairs if the probability of drawing a white chair is triple that of green chair determine the number of white chairs in the room.
Solution:
Let the number of white chairs = x
Given number of green chairs = 30
Total number of chairs in room = x + 30
Total outcomes in drawing a chair at random = x + 30
Number of outcomes favourable to green chair = 30
∴ P(G) = \(\frac{30}{x+30}\)
So, given in problem P(W) = 3 × \(\frac{30}{x+30}\)
= \(\frac{90}{x+30}\)
We know that P(G) + P(W) = 1
\(\frac{30}{x+30}\) + \(\frac{90}{x+30}\) = 1
\(\frac{120}{x+30}\) = 1
x + 30 = 120
x = 90
∴ No. of white chairs x = 90.

Question 3.
There are 25 cards of same size in a bag on which number 1 to 25 are written one card is taken out of the bag at random. Find the probability that the number on the selected card is not divisible by 5.
Solution:
Given total number of cards = 25
The number which are divisible by ‘5’ are 5, 10, 15, 20, 25
No. of all possible outcomes n(5) = 25
Number of out comes favourable to
E = n(E) = 5
∴ P(E) = \(\frac{n(E)}{n(S)}\)
= \(\frac{5}{25}\) = \(\frac{1}{5}\)
We know that P(E) + P(\(\overline{\mathrm{E}}\)) = 1
P(\(\overline{\mathrm{E}}\)) = 1 – (P(E))
p(\(\overline{\mathrm{E}}\)) = 1 – \(\frac{1}{5}\)
= \(\frac{5-1}{5}\)
P(\(\overline{\mathrm{E}}\)) = \(\frac{4}{5}\)
∴ probability that the number selected card is not divisible by 5 = \(\frac{4}{5}\)

Question 4.
A jar contains 18 marbles, some are red and other white if a marble is drawn at random from the jar the probability that it is white \(\frac{5}{6}\) . Find the number of white marbles.
Solution:
Total number of marbles in the jar = 18
Let the number of red marbles = k
The number of white marbles = 18 – k
probability of drawing a red marble = \(\frac{\mathrm{k}}{18}\)
From problem = \(\frac{\mathrm{k}}{18}\) = \(\frac{5}{6}\)
⇒ k = \(\frac{90}{6}\)
⇒ k = 15
No. of red marbles = k = 15
No. of white marbles = 18 – 15 = 3.

Question 5.
A game consists of tossing a one rupee coin 2 times and noting its outcome each time. Ravi wins if all the wins give the same result, i.e., two heads or two tails and lose otherwise. Calculate the probability that he will lose the game.
Solution:
We know if a coin is tossed for n times, then the total number of outcomes = 2ⁿ
So, a coin is tossed for 2 times, then the total number of outcomes 2² = 4.
see here
T T
T H
H T
H H
of the above no. of outcomes with different result = 2.
probability of lossing the game
= \(\frac{\text { No. of favourable outcomes of lose }}{\text { No. of total outcomes }}\)
= \(\frac{2}{4}\) = \(\frac{1}{2}\)

Question 6.
A lot consists of 200 ball pens of which 50 are defective and others are good. The shop keeper draws one pen at random and gives to sindhu. what is the probability that,
1) She will buy it ?
2) She will not buy it ?
Solution:
i) Total no. of ball pens = 200
∴ Number of all possible outcomes = 200
Number of defective ball pens = 50
Number of good ball pens = 200 – 50
⇒ No. of favourable outcomes = 150
probability that sindhu will buy it
= \(\frac{\text { No. of favourable outcomes }}{\text { Total No. of all possible outcomes }}\)
= \(\frac{150}{200}\) = \(\frac{3}{4}\)

ii) Probability that sindhu will not buy it
= 1 – (probability that sindhu will buy it)
= 1 – \(\frac{3}{4}\)
= \(\frac{4-3}{4}\)
= \(\frac{1}{4}\)

These TS 10th Class Maths Chapter Wise Important Questions Chapter 12 Applications of Trigonometry given here will help you to solve different types of questions.

TS 10th Class Maths Important Questions Chapter 12 Applications of Trigonometry

Previous Exams Questions

Question 1.
A person from the top of a building of height 25 m has observed another building top and bottom at an angle of elevation 45° and at an angle of depression 60° respectively. Draw a diagram for this data. (T.S. Mar. ’15)
Solution:

Question 2.
A ladder of 3.9 m length is laid against a wall. The distance between the foot of the wall and the ladder is 1.5 m. Find the height at which the ladder touches the wall. (T.S. Mar. ’15)
Solution:

h² = (3.9)² – (1.5)²
= (3.9 + 1.5) (3.9 – 1.5)
= 5. 4 × 2.4
= (0.6 × 9) × (0.6 × 4)
= (0.6)² × 6²
∴ h = 6 × 0.6 = 3.6 m

Question 3.
An observer flying in an aeroplane at an altitude of 900 m observes two ships in front of him, which are in the same direction at an angles of depression of 60° and 30° respectively. Find the distance between the two ships. (T.S. Mar. ’15)
Solution:

In ∆ABC
Tan 60 = \(\frac{900}{\mathrm{x}}\)
\(\sqrt{3}\) = \(\frac{900}{\mathrm{x}}\)
⇒ x = \(\frac{900}{\sqrt{3}}\) = 300\(\sqrt{3}\)
In ∆ ABD
Tan 30 = \(\frac{900}{\mathrm{x+d}}\)
\(\frac{1}{\sqrt{3}}\) = \(\frac{900}{300 \sqrt{3}+d}\)
d = 600\(\sqrt{3}\) m

Question 4.
If the angle of elevation of sun increases from ‘O’ to 90 then the length of shadow of a tower decreases. Is this true ? Justify your answer. (T.S. Mar. ’16)
Solution:
Yes, this statement is true.
We observe this in day to day life.

AD – ground
BC – tower
AB – shadow

Question 5.
A boat has to cross a river. It crosses river by making an angle of 60° with bank, due to the stream of river it travels a distance of 450 m to reach another side of river. Draw a diagram to this data. (T.S. Mar. ’15)
Solution:

AB – width of river
AD, BC are river banks
AC – The distance travelled in river = 450 m
A – initial point, C – terminal point

Question 6.
Two poles of equal heights are standing opposite to each other on either side of the road which is 80 m wide. From a point between them on the road, the angles of elevation of top of the poles are 60° and 30°. Find the height of poles, (T.S. Mar. ’16)
Solution:
As shown in the figure

AD = width of road = 80 m.
AB, DE are two poles
AB = DE (∵ they have equal heights)
‘C’ is a point on road.
∠ACB = 30°, ∠DCE = 60°
Then in ∆ ACB
tan C = \(\frac{\mathrm{AB}}{\mathrm{AC}}\)
⇒ tan 30 = \(\frac{\mathrm{AB}}{\mathrm{AC}}\)
\(\frac{1}{\sqrt{3}}\) = \(\frac{\mathrm{AB}}{\mathrm{AC}}\)
⇒ AC = AB\({\sqrt{3}}\) ……………….. (1)
In ∆ CDE
tan C = \(\frac{\mathrm{DE}}{\mathrm{CD}}\)
⇒ tan 60 = \(\frac{\mathrm{DE}}{\mathrm{CD}}\)
\({\sqrt{3}}\) = \(\frac{\mathrm{DE}}{\mathrm{CD}}\)
⇒ CD = \(\frac{\mathrm{DE}}{\sqrt{3}}\) ……………… (2)
but AC + CD = AD
⇒ AB\({\sqrt{3}}\) + \(\frac{\mathrm{DE}}{\sqrt{3}}\) = 80
But DE = AB
⇒ AB\({\sqrt{3}}\) + \(\frac{\mathrm{AB}}{\sqrt{3}}\) = 80
⇒ \(\frac{3 \mathrm{AB}+\mathrm{AB}}{\sqrt{3}}\) = 80
⇒ 4AB = 80\({\sqrt{3}}\)
⇒ AB = \(\frac{80 \sqrt{3}}{4}\) = 20 \({\sqrt{3}}\) m.
So height of the pole = 20 m.

Additional Questions

Question 1.
The angle of elevation of the top of a Tree from the foot of building is 60° and the angle of elevation of the top of the building from the foot of the tree is 30° what is the ratio of heights of tree and building.
Solution:

Let the height of the tree = x m = AB
Let the height of the building = y m = CD
distance between the tree and building = d = BD
Angle of elevation of the top of the tree = 60°
From the figure tan 60° = \(\frac{\mathrm{AB}}{\mathrm{BD}}\) = \(\frac{\mathrm{x}}{\mathrm{d}}\)
\(\sqrt{3}\) = \(\frac{\mathrm{x}}{\mathrm{d}}\)
d = \(\frac{\mathrm{x}}{\sqrt{3}}\) ……………. (1)
∴ tan 30° = \(\frac{\mathrm{CD}}{\mathrm{BD}}\) = \(\frac{\mathrm{y}}{\mathrm{d}}\)
⇒ \(\frac{1}{\sqrt{3}}\) = \(\frac{\mathrm{y}}{\mathrm{d}}\)
d = y\(\sqrt{3}\) ………………. (2)
\(\frac{\mathrm{x}}{\sqrt{3}}\) = y\(\sqrt{3}\)
\(\frac{\mathrm{x}}{\mathrm{y}}\) = \(\sqrt{3}\) × \(\sqrt{3}\)
\(\frac{\mathrm{x}}{\mathrm{y}}\) = \(\frac{3}{1}\)
x : y = 3 : 1
Hence, the ratio of the heights of the tree and building = 3 : 1.

Question 2.
The angle of the top of a pillar from two points are at a distance of 7m and 12m Find the height of the pillar from the base of the pillar and in the same straight line with it are complementary.
Solution:

From the figure,
Let AB be a height of the pillar = h m
Let the two points on the ground be ’c and d’ such that they make a distance 7 m and 12 m
From foot of the pillar
AC = 7m ; AD = 12m
Angles of elevation are ∠ACB = 0 ;
∠ADB = (90 – θ)
In the right angled ∆ABC, we have
tan θ = \(\frac{\mathrm{AB}}{\mathrm{AD}}\)
tan θ = \(\frac{\mathrm{h}}{7}\) …………… (1)
From.the right angled ∆ABC we have
tan (90 – θ) = \(\frac{\mathrm{AB}}{\mathrm{AD}}\)
cot θ = \(\frac{\mathrm{h}}{12}\)
\(\frac{1}{\tan \theta}\) = \(\frac{\mathrm{h}}{12}\)
tan θ = \(\frac{12}{\mathrm{~h}}\) …………….. (2)
From (1) and (2)
\(\frac{\mathrm{h}}{12}\) = \(\frac{12}{\mathrm{~h}}\)
h² = 84
h = \(\sqrt{4 \times 21}\)
h = 2 \(\sqrt{21}\) m
∴ The height of the pillar = h = 2 \(\sqrt{21}\)

Question 3.
A wire of length 25m 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.
Solution:

In the figure
Let PQ be a height of the electric pole = h m
given QS be the actual length of the wire = 25 m
let Q is length of the wire was cut S, R are the first and second points of observations.
Let PS = a + b ; PR = b
Angles of elevations are ∠PSQ = 30°
∠PRQ = 60°
From ∆PSQ
sin 30° = \(\frac{\mathrm{PQ}}{\mathrm{QS}}\)
\(\frac{1}{2}\) = \(\frac{\mathrm{PQ}}{\mathrm{QS}}\)
2 PQ = 25
PQ = 12.5
From ∆PRQ
tan 60° = \(\frac{\mathrm{PQ}}{\mathrm{QS}}\)
\(\sqrt{3}\) = \(\frac{12.5}{\mathrm{b}}\)
PR = b = \(\frac{12.5}{\sqrt{3}}\)
∆PQR, From
cos 60° = \(\frac{\mathrm{PR}}{\mathrm{QR}}\)
\(\frac{1}{2}\) = \(\frac{\frac{12.5}{\sqrt{3}}}{\mathrm{QR}}\)
\(\frac{1}{2}\) = \(\frac{12.5}{\sqrt{3 \mathrm{QR}}}\)
QR = \(\frac{25}{\sqrt{3}}\)
∴ Length of the wire was cut = \(\frac{25}{\sqrt{3}}\)

Question 4.
Two boys are on opposite sides of a tower 200m height. They measure the angle of elevation of the top of the tower as 45°, 60° respectively. Find the distance through which the boys are separated.
Solution:

Given height of Tower = 200 m
We say ‘x’ is the distance between 1^st person and base of the tower and ‘y’ is the distance between 2^nd person and base of the tower.
From ∆ ABD
Tan 45° = \(\frac{\mathrm{BD}}{\mathrm{AD}}\)
1 = \(\frac{200}{\mathrm{x}}\)
x = 200m
From ∆ BDC
Tan 60° = \(\frac{\mathrm{BD}}{\mathrm{DC}}\)

Question 5.
The tower height is l5mts and length of shadow is 15 \(\sqrt{3}\) m what is the angle of elevation of the sun.
Solution:

Let AC be the height of the tower = 15 m
Length of shadow = 15\(\sqrt{3}\) m
Let angle of elevation be θ.
From ∆ ABC Tan θ = \(\frac{\mathrm{AC}}{\mathrm{BC}}\) = \(\frac{15}{15 \sqrt{3}}\)
Tan θ = \(\frac{1}{\sqrt{3}}\)
Tan θ = Tan 30°
θ = 30°
∴ the angle of elevation of the sun θ = 30°