Class 10

Students can practice 10th Class Maths Study Material Telangana Chapter 10 Mensuration Ex 10.2 to get the best methods of solving problems.

TS 10th Class Maths Solutions Chapter 10 Mensuration Exercise 10.2

Question 1.
A toy is in the form of a cone mounted on a hemisphere. The diameter of the base and the height of the cone are 6 cm and 4 cm respectively. Determine the surface area of the toy. [use π = 3.14] (AS4)
Solution:
Diameter of the base of the cone (d)= 6 cm
∴ Radius of the base (r) = d/2 = 3 cm
Radius of the hemisphere = 3 cm
Let the slant height of the cone be ‘l’
l² = h² + r²
= 4² + 3²
= 16 + 9 = 25
∴ l = \(\sqrt{25}=\) = 5 cm.

Surface area of the toy = curved surface area of the cone + curved surface area of the hemisphere
= πrl + 2πr²
= \(\frac{22}{7}\) × 3 × 5 + 2 × \(\frac{22}{7}\) × 3 × 3
= \(\frac{22}{7}\) [3 × 5 + 2 × 3 × 3]
= \(\frac{22}{7}\) [15 + 18] = \(\frac{22}{7}\) × 33 = \(\frac{726}{7}\)
= 103.71 cm²

Question 2.
A Solid is in the form of a right circular cylinder with a hemisphere at one end and a cone at the other end. The radius of the common base is 8 cm. and heights of the cylindrical and conical portions are 10 cm and 6cm respectively. Find the total surface area of the solid. [use π = 3.14](AS4)
Solution:
Total surface area = C.S.A of the cone + C.S.A of cylinder + C.S.A of the hemisphere
Cone : Radius (r) = 8 cm
Height (h) = 6 cm
Slant height
l = \(\sqrt{\mathrm{r}^2+\mathrm{h}^2}\)
= \(\sqrt{8^2+6^2}\)
= \(\sqrt{64+36}\)
= \(\sqrt{100}\)
= 10 cm

C.S.A = πrl
= \(\frac{22}{7}\) × 8 × 10 = \(\frac{1760}{7}\) cm²
Cylinder : Radius (r) = 8 cm
Height(h) = 10 cm
C.S.A = 2πrh = 2 × \(\frac{22}{7}\) × 8 × 10
= \(\frac{3520}{7}\) cm
Hemisphere : Radius (r) = 8 cm
C.S.A = 2πr² = 2 × \(\frac{22}{7}\) × 8 × 8
= \(\frac{2816}{7}\) cm²
∴ Total surface area of the given solid
= \(\frac{1760}{7}\) + \(\frac{3520}{7}\) + \(\frac{2816}{7}\)
T.S.A = \(\frac{8096}{7}\) = 1156.57 cm²

Question 3.
A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends. The length of the capsule is 14 mm. and the width is 5mm. Find its surface area. (AS4)

Solution:
Surface area of the capsule = C.S.A of 2 hemispheres + C.S.A of the cylinder
Hemisphere : Radius (r) = \(\frac{\mathrm{d}}{2}\) = \(\frac{5}{2}\)
= 2.5 mm
C.S.A of two hemispheres = 2 × 2πr²
= 4 × \(\frac{22}{7}\) × 2.5 × 2.5 7
= \(\frac{550}{7}\) mm² = 78.57 mm²
Cylinder : Radius (r) = \(\frac{\mathrm{d}}{2}\) = \(\frac{5}{2}\) = 2.5 mm
Height (h) = 14 mm
C.S.A = 2πrh
= 2 × \(\frac{22}{7}\) × 2.5 × 14 = 220 mm²
∴ Surface area of the capsule = 78.57 + 220 = 298.57mm²

Question 4.
Two cubes each of volume 64 cm3 are joined end to end together. Find the surface area of the resulting cuboid. (AS1) (Mar ’15 (A.P.))
Solution:
Given, volume of the cube, V = a³ = 64 cm³
∴ a³ = 4 × 4 × 4 = 4³
Hence a = 4 cm
When two cubes are added, the length of cuboid = 2a
= 2 × 4 = 8 cm
breadth = a = 4 cm
height = a = 4 cm is formed

∴ T.S.A. of the cuboid
= 2 (lb + bh + lh)
= 2(8 × 4 + 4 × 4 + 8 × 4)
= 2(32 + 16 + 32)
= 2(80) = 160 cm².
∴ The surface area of resulting cuboid is 160 cm².

Question 5.
A storage tank consists of a circular cylinder with a hemisphere stuck on either end. If the external diameter of the cylinder be 1.4 in and its length be 8 m. Find the cost of pointing it on the outside at rate of ₹ 20 per m². (AS4)
Solution:
Total surface area of the tank = 2 × C.S.A of hemisphere + C.S.A of cylinder

Hemisphere :
Radius (r) = \(\frac{\text { diameter }}{2}\)
= \(\frac{1.4}{2}\)
= 0.7m
C.S.A of hemisphere = 2πr²
= 2 × \(\frac{22}{7}\) × 0.7 × 0.7 7
= 3.08 m²
2 × C.S.A = 2 × 3.08 m² = 6.16 m²
Cylinder : Radius (r) = \(\frac{\text { diameter }}{2}\)
= \(\frac{1.4}{2}\)
= 0.7 m
Height (h) = 8 m
C.S.A of the cylinder = 2πrh
= 2 × \(\frac{22}{7}\) × 0.7 × 8 = 35.2 m²
∴ Total surface area of the storage tank
= 35.2 + 6.16 = 41.36 m²
Cost of painting is surface area ₹ 20 per sq.m = 41.36 × 20 = ₹ 827.2

Question 6.
A sphere, a cylinder and a cone have the same radius and same height. Find the ratio of their volumes. (AS4)
[Hint: Diameter of the sphere is equal to heights of the cylinder and the cone].
Solution:
It is given that the three solids namely a sphere, a cylinder and a cone have the same radius. It is persumed that the three solids are of equal heights.

Sphere :
Radius of sphere = r
∴ volume of the sphere = \(\frac{4}{3}\) πr³.

Cylinder :
Radius of cylinder = r
Height of cylinder h = 2r (diameter of sphere)
∴ volume of cylinder = πr²h
= πr² × 2r
= 2πr × 2r
= 2πr³

Cone :
Radius of cone = r
Height of cone h = 2r
∴ Volume of cone = \(\frac{1}{3}\) πr²h
= \(\frac{1}{3}\) πr² × 2r
= \(\frac{2}{3}\) πr³
∴ The ratios of the volumes
= sphere : cylinder : cone
= \(\frac{4}{3}\) πr³ : 2πr³ : \(\frac{2}{3}\) πr³
= 4 πr³ : 6πr³ : 2πr³
= 4 : 6 : 2
= 2 : 3 : 1.

Question 7.
A hemisphere is cut out from one face of a cubical wooden block such that the diameter of the hemisphere is equal to the length of the cube. Determine the surface area of the remaining solid. (AS4)
Solution:
T.S.A. of the remaining solid = area of each surface of cube + Area of hemisphere – Area of cutting part of hemisphere.

Square surface : Side = a units.
Area a = a × a = a² sq. units.
6 × square surface area = 6a² sq. units.
Hemisphere : Diameter = a, Radius = \(\frac{\mathrm{a}}{2}\)
Area of hemisphere = 2πr²
= 2 × π × \(\left(\frac{\mathrm{a}}{2}\right)^2\)
= 2 × π × \(\frac{\mathrm{a}}{4}\) = \(\frac{\pi \mathrm{a}^2}{2}\)
Area of cutting part of hemisphere
= 2πr² = πr²
= \(\frac{\pi \mathrm{a}^2}{2}\) – π\(\left(\frac{a}{2}\right)^2\)
= \(\frac{\pi \mathrm{a}^2}{2}\) – \(\frac{\pi \mathrm{a}^2}{4}\) = \(\frac{\pi \mathrm{a}^2}{4}\)

Question 8.
A wooden article was made by scooping out a hemisphere from each end of a solid cylinder, as shown in the figure. If the height of the cylinder is 10 cm and its radius is of 3.5 cm, find the total surface are of the article. (AS4)
Solution:
Surface area of the given solid.
= C.S.A. of the cylinder + 2 × C.S.A. of hemisphre.

= 2πrh + 2 × 2πr²
= 2πrh + 4πr²
= 2 × \(\frac{22}{7}\) × 3.5 × 10 + 4 × \(\frac{22}{7}\) × 3.5 × 3.5
= 220 + 2(77)
= 220 + 154 = 374 cm²

Students can practice 10th Class Maths Study Material Telangana Chapter 11 Trigonometry Ex 11.2 to get the best methods of solving problems.

TS 10th Class Maths Solutions Chapter 11 Trigonometry Exercise 11.2

Question 1.
Evaluate the following. (AS₁)
i) sin 45° + cos 45°
Solution:
sin 45° + cos 45°
= \(\frac{1}{\sqrt{2}}\) + \(\frac{1}{\sqrt{2}}\)
= \(\frac{1+1}{\sqrt{2}}=\frac{2}{\sqrt{2}}=\frac{2}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}}=\frac{\sqrt{2}}{2}=\sqrt{2}\)

ii) \(\frac{\cos 45}{\sec 30 + cosec 60}\)
Solution:

iii) \(\frac{\sin 30+\tan 45-\ cosec 60}{\cot 45+\cos 60-\sec 30}\)
Solution:

iv) 2 tan² 45° + cos² 30° – sin² 60°
Solution:
2 tan² 45° + cos² 30° – sin² 60°
= 2(1)² + \(\left(\frac{\sqrt{3}}{2}\right)^2\) – \(\left(\frac{\sqrt{3}}{2}\right)^2\)
= \(\frac{2}{1}\) + \(\frac{3}{4}\) – \(\frac{3}{4}\)
= \(\frac{8+3-3}{4}\) = \(\frac{8}{4}\) = 2

v) \(\frac{\sec ^2 60-\tan ^2 60}{\sin ^2 30+\cos ^2 30}\)
Solution:

Question 2.
Choose the right option and justify your choice.
i) \(\frac{2 \tan 30}{1+\tan ^2 45}\)
(a) sin 60°
(b) cos 60°
(c) tan 30°
(d) sin 30°
Solution:
(c)

ii) \(\frac{1-\tan ^2 45}{1+\tan ^2 45}\)
(a) tan 90°
(b) 1
(c) sin 45°
(d) 0
Solution:
(d)
\(\frac{1-\tan ^2 45^{\circ}}{1+\tan ^2 45^{\circ}}=\frac{1-(1)^2}{1+(1)^2}=\frac{0}{1+1}=\frac{0}{2}\) = 0

iii) \(\frac{2 \tan 30}{1-\tan ^2 30}\)
(a) cos 60°
(b) sin 60°
(c) tan 60°
(d) sin 30°
Solution:
(c)

Question 3.
Evaluate sin 60° cos 30° + sin 30° cos 60°. What is the value of sin (60° + 30°) ? What can you conclude ? (AS₁, AS₃)
Solution:
Take sin 60°. cos 30° + sin 30°. cos 60°
= \(\frac{\sqrt{3}}{2}\) . \(\frac{\sqrt{3}}{2}\) + \(\frac{1}{2}\) . \(\frac{1}{2}\)
= \(\frac{(\sqrt{3})^2}{4}\) + \(\frac{1}{4}\) = \(\frac{3}{4}\) + \(\frac{1}{4}\)
= \(\frac{3+1}{4}\) = \(\frac{4}{4}\) = 1 …………….. (1)
Now take sin (60° + 30°) = sin 90° = 1 ………….. (2)
From equations (1) and (2), I conclude that sin (60° + 30°) = sin 60°. cos 30° + sin 30° . cos 60°
i.e., sin (A + B) = sin A . cos B + cos A . sin B

Question 4.
Is it right to say cos (60° + 30°) = cos 60°. cos 30° – sin 60°. sin 30° ? (AS₂, AS₃)
Solution:
L.H.S. = cos (60° + 30°)
= cos 90° = 0
R.H.S. = cos 60° . cos 30° – sin 60° . sin 30°
= \(\frac{1}{2}\) . \(\frac{\sqrt{3}}{2}\) – \(\frac{\sqrt{3}}{2}\) . \(\frac{1}{2}\)
= \(\frac{\sqrt{3}}{4}\) – \(\frac{\sqrt{3}}{4}\) = 0
∴ L.H.S. = R.H.S.
Yes, it is right to say
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 5.
In right angled triangle ∆PQR, right angle is at Q and PQ = 6 cm, ∠RPQ = 60°. Determine the lengths of QR and PR. (AS₄)
Solution:
In ∆PQR, ∠Q = 90°
∠RPQ = 60°

Question 6.
In ∆XYZ, right angle is at Y, YZ = x and XY = 2x, then determine ∠YXZ and ∠YZX. (AS₄)
Solution:
In ∆XYZ, ∠YXZ = ?
Given that ∠XYZ = 90°
sin x = \(\frac{Y Z}{X Y}=\frac{X}{2 x}=\frac{1}{2}\)
We know that sin 30° = \(\frac{1}{2}\)
∴ x = 30°
In ∆XYZ, ∠Y = 90°, ∠X = 30°
∠YZX = 180° – (∠Y + ∠X)

= 180° – (90° + 30°)
= 180° – 120°
= 60°

Question 7.
Is it right to say that sin (A + B) = sin A + sin B ? Justify your answer. (AS₂)
Solution:
Take A = 60°, B = 30°
Then sin (A + B) = sin (60° + 30°)
= sin 90° = 1
sin A + sin B = sin 60° + sin 30°
= \(\frac{\sqrt{3}}{2}\) + \(\frac{1}{2}\)
= \(\frac{\sqrt{3}+1}{2}\)
∴ sin (A + B) ≠ sin A + sin B
It is not right to say that sin (A + B) = sin A + sin B

Students can practice 10th Class Maths Study Material Telangana Chapter 11 Trigonometry Ex 11.1 to get the best methods of solving problems.

TS 10th Class Maths Solutions Chapter 11 Trigonometry Exercise 11.1

Question 1.
In a right angle triangle ABC, 8 cm, 15 cm and 17 cm are the lengths of AB, BC and CA respectively. Then, find out sin A, cos A and tan A. (AS₁)
Solution:
Given that ∆ABC, AB = 8 cm; BC = 15 cm; CA = 17 cm

Question 2.
The sides of a right angle triangle PQR are PQ = 7 cm, QR = 25 cm and ∠Q = 90° respectively. Then find tan Q – tan R.
(AS₁)
Solution:
Given that in ∆PQR, PQ = 7 cm
QR = 25 cm and RP = 24 cm
PR = \(\sqrt{\mathrm{QR}^2-\mathrm{PQ}^2}\) = \(\sqrt{25^2-7^2}\)
= \(\sqrt{625-49}\) = \(\sqrt{576}\) – 24
In the text given problem is wrong. We take
∠P = 90° instead of ∠Q = 90°
tan P = \(\frac{\text { Side opposite to } \angle \mathrm{P}}{\text { Side adjacent to } \angle \mathrm{P}}\)
= \(\frac{\mathrm{PR}}{\mathrm{PQ}}\) = \(\frac{24}{7}\)
tan R = \(\frac{\text { Side opposite to } \angle \mathrm{R}}{\text { Side adjacent to } \angle \mathrm{R}}\)
= \(\frac{\mathrm{PQ}}{\mathrm{PR}}\) = \(\frac{7}{24}\)

∴ tan P – tan R = \(\frac{24}{7}\) – \(\frac{7}{24}\)
= \(\frac{576-49}{168}\) = \(\frac{27}{168/}\)

Question 3.
In a right tingle triangle ABC with right angle at B, in which a = 24 units, b = 25 units and ∠BAC = θ. Then, find cos θ and tan θ.
Solution:
In ∆ABC, ∠B = 90°
a = BC = 24 units
b = AC = 25 units
In ∆ABC,
∴ AC² = AB² + BC²
∠B = 90° (Pythagoras theorem)

25² = AB² + 24²
⇒ AB² = 25² – 24²
= 625 – 576 = 49
⇒ AB = \(\sqrt{49}\) = 7
c = AB = 7 units
cos θ = \(\frac{\text { Side adjacent to } \theta}{\text { Hypotenuse }}\)
= \(\frac{\mathrm{AB}}{\mathrm{AC}}\) = \(\frac{7}{25}\)
sin θ = \(\frac{\text { Side opposite to } \theta}{\text { Hypotenuse }}\)
= \(\frac{\mathrm{BC}}{\mathrm{AC}}\) = \(\frac{24}{25}\)
∴ tan θ
= \(\frac{\sin \theta}{\cos \theta}=\frac{\mathrm{BC}}{\mathrm{AC}} \times \frac{\mathrm{AC}}{\mathrm{AB}}=\frac{24}{25} \times \frac{25}{7}=\frac{24}{7}\)

Question 4.
If cos A = \(\frac{12}{13}\), then find sin A and tan A. (AS₁) (A.P. Mar. ’16)
Solution:
Given that, cos A = \(\frac{12}{13}\)
and cos A = \(\frac{\text { Adjacent side }}{\text { Hypotenuse }}\)
For angle A, adjacent side = AB = 12 k Hypotenuse = AC = 13 k (Where k is a positive number)

Now, we have in ∆ABC
AC² = AB² + BC² (Pythagoras theorem)
⇒ (13k)² = (12k)² + BC²
⇒ BC² = (13k)² – (12k)²
= 169 k² – 144 k²
= 25 k²
∴ BC = \(\sqrt{25 \mathrm{k}^2}\) = 5 k
BC = 5 k = Opposite side

Question 5.
If 3 tan A = 4, then find sin A and cos A. (AS₁)
Solution:
Given that, 3 tan A = 4

Opposite side to ∠A = BC = 4k
Adjacent side to ∠A = AB = 3k
Now we have in ∆ABC, ∠B = 90°
∴ AC² = AB² + BC² (By Pythagoras theorem)
= (3k)² + (4k)²
= 9k² + 16k²
= 25k²
∴ AC = \(\sqrt{25 \mathrm{k}^2}\) = 5k
AC = 5k = Hypotenuse
sin A = \(\frac{\text { Opposite side }}{\text { Hypotenuse }}\)
= \(\frac{4 \mathrm{k}}{5 \mathrm{k}}\) = \(\frac{4}{5}\)
cos A = \(\frac{\text { Adjacent side }}{\text { Hypotenuse }}\)
= \(\frac{3 \mathrm{k}}{5 \mathrm{k}}\) = \(\frac{4}{5}\)

Question 6.
If ∠A and ∠X are acute angles such that cos A = cos X, then show that ∠A = ∠X. (AS₂)
Solution:
In the given triangle,

⇒ \(\frac{\mathrm{AC}}{\mathrm{AX}}\) = \(\frac{\mathrm{XC}}{\mathrm{AX}}\)
⇒ AC = XC
⇒ ∠A = ∠X
(∴ Angles opposite to equal sides are also equal)

Question 7.
Given cot θ = \(\frac{7}{8}\), then evaluate
i) \(\frac{(1+\sin \theta)(1-\sin \theta)}{(1+\cos \theta)(1-\cos \theta)}\)
ii) \(\frac{(1+\sin \theta)}{\cos \theta}\) (AS₁)
Solution:
Given,

Let AB = 7k and BC = 8k
In a right angled triangle,
AC² = AB² + BC² (By Pythagoras theorem)
= (7k)² + (8k)²
= 49 k² + 64 k²
AC² = 113 k²
AC = \(\sqrt{113}\) k
Now,

Question 8.
In a right angle triangle ABC, right angle Is at B, If tan A = \(\sqrt{3}\), then find the value of
i) sin A cos C + cos A sin C
ii) cos A cos C – sin A sin C (AS₁)
Solution:
Given, tan A = \(\frac{\sqrt{3}}{1}\)

Let opposite side = \(\sqrt{3}\)k and adjacent side = 1 k
In right angled ∆ABC
AC² = AB² + BC² (By Pythagoras theorem)
⇒ AC² = (1k)² + (\(\sqrt{3}\)k)²
⇒ AC² = 1k² + 3k²
⇒ AC² = 4k² ⇒ AC = \(\sqrt{4 \mathrm{k}^2}\)
AC = 2k
Now,
sin A = \(\frac{\mathrm{BC}}{\mathrm{AC}}=\frac{\sqrt{3} \mathrm{k}}{2 \mathrm{k}}=\frac{\sqrt{3}}{2}\)
cos A = \(\frac{\mathrm{AB}}{\mathrm{AC}}=\frac{1 \mathrm{k}}{2 \mathrm{k}}=\frac{1}{2}\)

ii) cos A . cos C – sin A . sin C
= \(\frac{1}{2}\) . \(\frac{\sqrt{3}}{2}\) – \(\frac{\sqrt{3}}{2}\) . \(\frac{1}{2}\)
= \(\frac{\sqrt{3}}{4}\) – \(\frac{\sqrt{3}}{4}\) = 0

Students can practice 10th Class Maths Solutions Telangana Chapter 4 Pair of Linear Equations in Two Variables Ex 4.2 to get the best methods of solving problems.

TS 10th Class Maths Solutions Chapter 4 Pair of Linear Equations in Two Variables Exercise 4.2

Question 1.
The ratio of incomes of two persons is 9: 7 and the ratio of their expenditures is 4 : 3. If each of them manages to save
₹ 2,000 per month, find their monthly income.
Solution:
Let the income of the first person be ₹ 9x and that of the second person be ₹ 7x.
Further, let the expenditure of the first person and the second person be ₹ 4y and 3y respectively.
Then, the saving of first person = 9x – 4y
By problem, 9x – 4y = 2,000 —— (1)
The saving of second person = 7x – 3y
Solving (1) and (2), we get
We equate the coefficients of ‘y’ in the equations.

∴ x = 2,000
Substitute x = 2,000 in equation (1), we get
9 × 2,000 – 4y = 2,000
18,000 – 4y = 2,000
-4y = 2,000 – 18,000 = -16,000
∴ y = \(\frac{-16000}{-4}\) = 4,000
Therefore, monthly income of the first person = ?9x
= ₹ 9 × 2,000
= ₹ 18,000
Monthly income of the second person = ₹ 7x
= ₹ 7 × 2,000
= ₹ 14,000

Question 2.
The sum of a two digit number and the number obtained by reversing the digits is 66. If the digits of the number differ by 2, find the number. How many such numbers are there ?
Solution:
Let the digit at units place be x.
And the digit at tens place be y.
The number will be yx.
The value of the number = y × 10 + x × 1
= 10y + x

Number obtained by reversing the digits = xy
Then the value of the reversed number = 10 × x + y × 1
= 10x + y
By problem, we have
(10y + x) + (10x + y) = 66
⇒ 11x + 11y = 66
⇒ x + y = 6 —– (1)
It is given that the digits differ by 2.
So, x – y = 2 (or) y – x = 2 —- (2)
Solving (1) and (2), we get

Substitute x = 4 in (1), we get
x + y = 6 ⇒ y = 6 – 4 = 2
Substitute y = 4 in (1),
x + y = 6 ⇒ x = 6 – 4 = 2
∴ The required number is 24.
∴ The required number is 42.
There are two numbers (i.e.,) 24 and 42.

Question 3.
The larger of two supplementary angles exceeds the smaller by 18°. Find the angles.
Solution:
Two angles are said to be supplementary if their sum is 180°.
Let the smaller supplementary angle be x°. and the larger supplementary angle be y°.
We know that sum of these two angles is 180°.
∴ x + y = 180° —- (1)
The larger angle exceeds the smaller by 18°. Then, y = x + 18
⇒ -x + y = 18 —- (2)
Solving (1) & (2), we get

Substitute y = 99 in (1), we get
x + 99 = 180
⇒ x = 180 – 99 = 81
Therefore, the required angles are 81°, 99°.

Question 4.
The taxi charges in Hyderabad are fixed along with the charge for the distance covered. For a distance of 10 km., the charge paid is ₹ 220. For a journey of 15 km, the charge paid is ₹ 310.
i) What are the fixed charges and charge per km ?
ii) How much does a penon have to pay for travelling a distance of 25 km?
Solution:
Let the fixed charge be ₹ x.
And charge per km be ₹y.
For a distance of 10km, the charge paid is ₹ 220.
Then x + 10y = 220 —– (1)
For a distance of 15 km, the charge paid is ₹ 310.
Then, x + 15y = 310 —- (2) ;
Solving (1) & (2), we get

Substitute y = 18 in equation (1), we get
x + (10 × 18) = 220
x + 180 = 220
∴ x = 220 – 180 = 40
Fixed charge = ₹ 40
Charge per km = ₹ 18
∴ Charge for 25 km = ₹ 450

Question 5.
A fraction becomes if \(\frac{4}{5}\) is added to both numerator and denominator. If, however, 5 is subtracted from both numerator and denominator, the fraction becomes \(\frac{1}{2}\). What is the fraction ? (AP Mar. 15)
Solution:
Let the fraction be; \(\frac{x}{y}\)
If 1 is added to both numerator and denominator then the fraction = \(\frac{x+1}{y+1}\)
By problem, \(\frac{x+1}{y+1}\) = \(\frac{4}{5}\)
⇒ 5(x + 1)= 4(y + 1)
⇒ 5x + 5 = 4y +4
⇒ 5x – 4y = 4 – 5 = -1
⇒ 5x – 4y = -1 —- (1)
If 5 is subtracted from both numerator and denominator, then
the fraction = \(\frac{x-5}{y-5}\)
By problem, \(\frac{x-5}{y-5}\) = \(\frac{1}{2}\)
⇒ 2(x – 5) = 1(y – 5)
⇒ 2x – 10 = y – 5
⇒ 2x – y = -5 + 10 = 5
⇒ 2x – y = 5 —- (2)
Solving (1) & (2), we get

Substitute x = 7 in (2),
⇒ 2 × 7 – y = 5
⇒ 14 – y = 5
⇒ -y = 5 – 14 = -9
∴ y = 9
∴ The required fraction is \(\frac{7}{9}\).

Question 6.
Places A and B are 1oo km apart on a highway. One car starts from A and another from B at the same time at different speeds. if the cars travel in the same direction, they meet in 5 hours. If they travel towards each other, they meet in 1 hour. What are the speeds of the two cars?
Solution:
Let the speed of the car that starts from A be x kmph.
Let the speed of the car that starts from B be y kmph.
Distance between A and B is 1oo km.

∴ Speed × Time = Distance travelled
(x – y) × 5 = 100
5x – 5y = 100
⇒ x – y = 20 —- (1)
If the two cars travel in the opposite direction, their relative speed will be (x + y) kmph. It is also given that the cars will meet in 1 hour.
∴ Speed × Time = Distance travelled
(x + y) × 1 = 100
⇒ x + y = 100 —- (2)
Solving (1) & (2), we get

Substitute x = 60 in equation (2),
60 + y = 100
∴ y = 100 – 60 = 40
∴ The speeds of the cars are 60 kmph and 40 kmph.

Question 7.
Two angles are complementary. The larger angle is 3° less than twice the measure of the smaller angle. Find the
measure of each angle.
Solution:
Two angles are said to be complementary if their sum is 90°.
Let the smaller angle be x° and the larger angle y°.
∴ x + y = 90° —- (1)
Given that the larger angle is 30 less than twice the measure of the smaller angle.
∴ y = 2x – 3
⇒ 2x – y = 3 — (2)
Solving (1) & (2), we get

Substitute x = 31 in (1),
Then, 31 + y = 90
∴ y = 90 – 31 = 59
The two angles are 31° and 59°

Question 8.
An algebra textbook has a total of 1382 pages. It is broken up into two parts. The second part of the book has 64 pages
more than the first part. How many pages are in each part of the book ?
Solution:
Let the number of pages in the first part of algebra textbook be ‘x’ and that of second part be ‘y’. The textbook contains 1382 pages.
∴ x + y = 1382 —- (1)
It is given that the second part of the book has 64 pages more than the first part.
∴ y = x + 64
⇒ x – y = – 64 —- (2)
Solving (1) & (2), we get

Substitute x = 659 in (1), we get
659 + y = 1382
∴ y = 1382 – 659 = 723
First part of the book has 659 pages whereas the second part has 723 pages.

Question 9.
A chemist has two solutions of hydrochloric acid in stock. One is 50% solution and the other is 80% solution. How much of each should be used to obtain 100 ml of a 68% solution ?
Solution:
Let 50% solution to be used be ‘x’ ml.
And 80% solution to be used be ‘y’ ml.
Quantity of solution to be obtained = 100 ml
∴ x + y = 100 —– (1)
50% of ‘x’ ml = \(\frac{x \times 50}{100}\) = \(\frac{1}{2} \mathrm{x}\)
80% of ‘y’ ml = \(\frac{\mathrm{y} \times 80}{100}\) = \(\frac{4}{5} \mathrm{y}\)
By problem, \(\frac{1}{2}\)x + \(\frac{4}{5}\)y = 68
Multiplying each terms by 10, we get
[\(\frac{1}{2}\)x × 10] + [\(\frac{4}{5}\)y × 10] = 68 × 10
5x + 8y = 680 —– (2)
Solving equation (1) & (2), we have

∴ y = \(\frac{180}{3}\) = 60
Substitute y = 60 in (1), we have
x + y = 100
x + 60 = 100
∴ x = 100 – 60 = 40
Therefore, 50% solution to be used = 40 ml.
80% solution to be used 60 ml

Question 10.
Suppose you have ₹ 12,000 to invest. You have to invest some amount at 10% and the rest at 15%. How much should be
invested at each rate to yield 12% on the total amount invested?
Solution:
Total amount to be invested = ₹ 12,000/-
Let the amount invested at 10% be ₹ x.
Let the amount invested at 15% be ₹ y.
∴ x + y = ₹ 12,000 —- (1)
Amount yielded at 10% of
x = \(\frac{x \times 10}{100}\) = \(\frac{x}{10}\)
Amount yielded at 15% of
y = \(\frac{y \times 15}{100}\) = \(\frac{3 \mathrm{y}}{20}\)
yielded at 12% of ₹ 12,000
= \(\frac{12000 \times 12}{100}\) = 1440
∴ \(\frac{x}{10}\) + \(\frac{3 y}{20}\) = \(\frac{1440}{1}\)
∴ 2x + 3y = 28800 —- (2)

Substitute y = 4800 in (1), we get
x + 4800 = 12000
∴ x = 12000 – 4800 = 7200
The amounts to be invested at 10% and 15% are ₹ 7,200 and ₹ 4,800.

Students can practice 10th Class Maths Solutions Telangana Chapter 4 Pair of Linear Equations in Two Variables InText Questions to get the best methods of solving problems.

TS 10th Class Maths Solutions Chapter 4 Pair of Linear Equations in Two Variables InText Questions

Think – Discuss

Question 1.
Two situations are given below. (Page No. 73)
i) The cost of 1 kg potatoes and 2 kg tomatoes was ₹ 30 on a certain day. After two days, the cost of 2 kg potatoes and 4 kg tomatoes was found to be ₹ 66.
ii) The coach of a cricket team of M.K. Nagar High School buys 3 bats and 6 balls for ₹ 3900. Later he buys one more bat and 2 balls for ₹ 1300.
Identify the unknowns in each situation. We observe that there are two unknowns in each case.
Solution:
i) Let the cost of 1 kg potatoes be ₹ x.
The cost of 1 kg tomatoes be ₹ y.
Given the cost of 1 kg potatoes and 2 kg tomatoes = ₹ 30
∴ x + 2y = 30
Given the cost of 2 kg potatoes and 4 kg tomatoes = ₹ 66
∴ 2x + 4y = 66
The pair of linear equations in two variables are x + 2y = 30 and 2x + 4y = 66.
We observe that there are two unknowns in this case.

ii) Let the cost of one bat be ₹ x.
The cost of one ball be ₹ y.
Given the cost of 3 bats and 6 balls = ₹ 3,900
3x + 6y = 3,900
Given 4 bats and 2 balls cost = ₹ 1,300
∴ 4x + 2y = 1,300
The pair of linear equations in two variables are 3x + 6y = 3,900 and 4x + 2y = 1300
We observe that there are two unknowns in this case.

Question 2.
Is a dependent pair of linear equations always consistent ? Why or why not ? (Page No. 79)
Answer:
If the lines intersect at a point gives the unique solution of the equations.
If the lines coincide then there are infinitely many solutions each point on the line being a solution. No, a dependent pair of linear equations are always consistent.

Try This

Mark the correct option in the following questions :

Question 1.
Which of the following equations is not a linear equation ?
a) 5 + 4x = y + 3
b) x + 2y = y – x
c) 3 – x = y² + 4
d) x + y = 0 (Page No. 75)
Solution:
a) 5 + 4x = y + 3
4x + 5 = y + 3
4x + 5 – y – 3 = 0
4x – y + 2 = 0
This equation is in the form of ax + by + c = 0 where a, b, c are real numbers.
So, the given equation is a linear equation.

b) x + 2y = y – x
x + 2y – y + x = 0
2x + y = 0
This equation is in the form of ax + by + c = 0 where a, b, c are real numbers.
So, the given equation is a linear equation.

c) 3 – x = y² + 4
3 – x – y² – 4 = 0
– x – y² – 1 = 0
x + y² + 1 = 0
This equation is not in the form of ax + by + c = 0 where a, b, c are real numbers.
∴ The given equation is not a linear equation.

d) x + y = 0
This equation is in the form of ax + by + c = 0 where a, b, c are real numbers.
So, the given equation is a linear equation.

Question 2.
Which of the following is a linear equation in one variable ? ( b )
a) 2x + 1 = y – 3
b) 2t – 1 = 2t + 5
c) 2x – 1 = x²
d) x² – x + 1 = 0 (Page No. 76)
Solution:
a) 2x + 1 = y – 3
⇒ 2x + 1 – y + 3 = 0
⇒ 2x – y + 4 = 0
Is a linear equation in two variables.
They are x and y.

b) 2t – 1 = 2t + 5
⇒ 2t – 1 – 2t – 5 = 0
⇒ -6 = 0 (False)
Is not a linear equation in one variable.

c) 2x – 1 = x² ⇒ x² – 2x + 1 = 0
Is a linear equation in one variable i.e., ‘x’.

d) x² – x + 1 = 0
Is a linear equation in one variable.
i.e., ‘x’.

Question 3.
Which of the following numbers is a solution for the equation
2(x + 3) = 18 ? ( b )
a) 5
b) 6
c) 13
d) 21 (Page No. 76)
Solution:
Given equation 2(x + 3) = 18
a) At x = 5; 2 (5 + 3) = 18
2 × 8 = 18
16 = 18 (False) Not a solution.

b) At x = 6; 2(6 + 3) = 18
2 × 9 = 18
18 = 18 (True)
x = 6 is a solution for the given equation.

c) At x = 13; 2(13 + 3) = 18
2 × 16 = 18
32 = 18 (False) Not a solution.

d) At x = 21; 2(21 + 3) = 18
2 × 24 = 18
48 = 18 (False) Not a solution.

Question 4.
The value of x which satisfies the equation 2x – (4 – x) = 5 – x is
a) 4.5
b) 3
c) 2.25
d) 0.5 ( c ) (Page No. 76)
Solution:
Given equation 2x – (4 – x) = 5 – x
a) At x = 4.5; 2 (4.5) – (4 – 4.5) = 5 – 4.5
9 – (-0.5) = (0.5)
9 + 0.5 = 0.5
9.5 = 0.5 (False)
∴ Value of x does not satisfies the equation.

b) At x = 3 ; 2(3) – (4 – 3) = 5 – 3
6 – 1 = 2
5 = 2 (False)
∴ Value of x does not satisfies the equation.

c) At x = 2.25 ; 2(2.25) – (4 – 2.25)
= 5 – 2.25
4.50-1.75 = 2.75
2.75 = 2.75 (True)
∴ Value of x satisfies the equation.

d) At x = 0.5 ; 2(0.5) – (4 – 0.5) = 5 – 0.5
1 – 3.5 = 4.5
– 2.5 = 4.5 (False)
∴ Value of x does not satisfies the equation.

Question 5.
The equation x – 4y = 5 has
a) no solution
b) unique solution
c) two solutions
d) infinitely many solutions (d) (Page No. 76)
Solution:
Only one equation with two unknowns (variables) we can find many solutions.

Question 6.
In the example above can you find the cost of each bat and ball ? (Page No. 79)
Solution:
No, we cannot find the cost of each bat and ball because the equations are geometrically shown by a pair of coincident lines. Every point on the line is a common solution to both the equations.

Question 7.
For what value of ‘p’ for the following pair of equations has a unique solution.
2x + py = -5 and 3x + 3y = -6 (Page No. 83)
Solution:
Given equations are 2x + py = -5 and 3x + 3y = -6
a₁ = 2 ; b₁ = p; c₁ = -5
a₂ = 3; b₂ = 3 ; c₂ = -6
\(\frac{\mathrm{a}_1}{\mathrm{a}_2}\) = \(\frac{2}{3}\) ; \(\frac{\mathrm{b}_1}{\mathrm{b}_2}\) = \(\frac{p}{3}\) ; \(\frac{\mathrm{c}_1}{\mathrm{c}_2}\) = \(\frac{-5}{-6}\)
Given the pair of equations has a unique solution.
∴ \(\frac{\mathrm{a}_1}{\mathrm{a}_2}\) ≠ \(\frac{\mathrm{b}_1}{\mathrm{b}_2}\) (∵ p ≠ 2)
If p = except 2 then the given pair of equations has a unique solution.

Question 8.
Find the value of ‘k’ for which the pair of equations 2x – ky + 3 = 0, 4x + 6y – 5 = 0 represent parallel lines. (Page No. 83)
Answer:
Given pair of equations
2x – ky + 3 = 0 and 4x + 6y – 5 = 0
a₁ = 2; b₁ = -k; c₁ = 3
a₂ = 4; b₂ = 6; c₂ = -5
Given the pair of lines are parallel.
∴ \(\frac{a_1}{a_2}\) = \(\frac{b_1}{b_2}\) ≠ \(\frac{c_1}{c_2}\)
\(\frac{a_1}{a_2}\) = \(\frac{\mathrm{b}_1}{\mathrm{~b}_2}\) ⇒ \(\frac{2}{4}\) = \(\frac{-\mathrm{k}}{6}\)
-4k = 2 × 6
-4k = 12
∴ k = \(\frac{12}{-4}\) = -3

Question 9.
For what value of ‘k’ for the pair of equations 3x + 4y + 2 = 0 and 9x + 12y + k = 0 represent coincident lines. (Page No. 83)
Solution:
Given pair of equations 3x + 4y + 2 = 0 and 9x + 12y + k = 0 (A.P.Mar.’16) (A.P. Jun.’15)
Given the pair of lines are coincident.
∴ \(\frac{a_1}{a_2}\) = \(\frac{b_1}{b_2}\) = \(\frac{c_1}{c_2}\)
\(\frac{a_1}{a_2}\) = \(\frac{c_1}{c_2}\) ⇒ \(\frac{3}{9}\) = \(\frac{2}{\mathrm{k}}\) ⇒ \frac{1}{3}\(\) = \(\frac{2}{k}\)
∴ k = 3 × 2 = 6

Question 10.
For what Positive values of ‘p’ the following pair of linear equations have infinitely many solutions ? (Page No. 83)
px + 3y – (p – 3) = 0
12x + py – p = 0
Solution:
Given pair of equations are px + 3y – (p – 3) = 0 and 12x + py – p = 0
a₁ = p; b₁ = 3; c₁ = -(p – 3);
a₂ = 12; b₂ = p; c₂ = -p
Given equations has infinitely many solutions.
∴ \(\frac{\mathrm{a}_1}{\mathrm{a}_2}\) = \(\frac{\mathrm{a}_1}{\mathrm{a}_2}\) = \(\frac{\mathrm{c}_1}{\mathrm{c}_2}\)
⇒ \(\frac{\mathrm{p}}{12}\) = \(\frac{3}{p}\) = \(\frac{-(p-3)}{-p}\)
⇒ \(\frac{\mathrm{p}}{12}\) = \(\frac{3}{p}\) = \(\frac{-\mathrm{p}+3}{-\mathrm{p}}\) ⇒ \(\frac{\mathrm{p}}{12}\) = \(\frac{3}{p}\) = \(\frac{\mathrm{p}-3}{\mathrm{p}}\)
⇒ p × p = 3 × 12
⇒ p² = 36
⇒ p = \(\sqrt{36}\) = 6

Do This

Question 1.
Solve the following systems of equations : (Page No. 79)

i) x – 2y = 0
3x + 4y = 20
Solution:
i) x – 2y = 0
-2y = -x
y = \(\frac{x}{2}\)

3x+ 4y = 20
4y = 20 – 3x
y = \(\frac{20-3 x}{4}\)

Scale :X-axis – 1 unit = 1 cm
Y – axis – 1 unit = 1 cm

The two lines are intersecting lines, meet at (4, 2).
The solution set is {(4, 2)}.

ii)
x + y = 2
2x + 2y = 4
Solution:
x + y = 2
y = 2 – x

2x + 2y = 4
2y = 4 – 2x
y = \(\frac{4-2 x}{2}\)

Scale : X – axis – 1 unit = 1 cm
Y – axis – 1 unit = 1 cm

Scale : X – axis – 1 unit = 1 cm
Y – axis – 1 unit = 1 cm
These two lines are coincident lines.
∴ There are infinitely many solutions.

iii) 2x – y = 4; 4x – 2y = 6
Solution:
2x – y = 4 ⇒ y = 2x – 4

4x – 2y = 6 ⇒ 2y = 4x – 6
⇒ y = 2x – 3

These two are parallel lines.
∴ The pair of linear equations has no solution

Scale: X – axis – 1 unit = 1 cm
Y – axis – 1 unit = 1 cm

Question 2.
Two rails on a railway track are represented by the equations.
x + 2y – 4 = 0 and 2x + 4y – 12 = 0.
Represent this situation graphically. (Page No. 79)
Solution:
x + 2y – 4 = 0; 2y = 4 – x
y = \(\frac{4-x}{2}\)
x + 2y – 4 = 0

2x + 4y – 12 = 0
⇒ 4y = 12 – 2x ⇒ 4y = 2(6 – x)
⇒ y = \(\frac{6-x}{2}\)

∴ These lines are parallel and hence no solution.

Question 3.
Check each of the given systems of equations to see if It has a unique solution, infinitely many solutions or no solution. Solve them graphically. (A.P. Mar.16’) (Page No. 83)

i) 2x + 3y = 1
3x – y = 7
Solution:
Let a₁x + b₁y + c₁ = 0 \(\simeq\) 2x + 3y – 1 = 0
a₂x + b₂y + c₂ = 0 \(\simeq\) 3x – y – 7 = 0
Now comparing their coefficients i.e.,
\(\frac{a_1}{a_2}\) and \(\frac{b_1}{b_2}\) ⇒ \(\frac{2}{3}\) ≠ \(\frac{3}{-1}\)
∴ The given lines are intersecting lines.


Scale: X – axis – 1 unit = 1 cm
Y – axis – 1 unit = 1 cm
2x + 3y = 1
⇒ 3y = 1 – 2x
⇒ y = \(\frac{1-2 x}{3}\)
∴ The system of equation has a unique solution (2, -1).

ii) x + 2y = 6
2x + 4y = 12
Solution:
From the given pair of equations,
\(\frac{\mathrm{a}_1}{\mathrm{a}_2}\) = \(\frac{1}{2}\) ; \(\frac{\mathrm{b}_1}{\mathrm{b}_2}\) = \(\frac{2}{4}\) ; \(\frac{\mathrm{c}_1}{\mathrm{c}_2}\) = \(\frac{6}{12}\) = \(\frac{1}{2}\)
∴ \(\frac{\mathrm{a}_1}{\mathrm{a}_2}\) = \(\frac{\mathrm{b}_1}{\mathrm{b}_2}\) = \(\frac{\mathrm{c}_1}{\mathrm{c}_2}\)
∴ The lines are dependent and have infinitely many solutions.
x + 2y = 6 ⇒ 2y = 6 – x
⇒ y = \(\frac{6-x}{2}\)

2x + 4y = 12
⇒ 4y = 12 – 2x ⇒ y = \(\frac{12-2 \mathrm{x}}{4}\)


Scale: X – axis – 1 unit = 1 cm
Y – axis – 1 unit = 1 cm

iii) 3x + 2y = 6
6x + 4y = 18
Solution:
From the given equations
\(\frac{\mathrm{a}_1}{\mathrm{a}_2}\) = \(\frac{3}{6}\) = \(\frac{1}{2}\) ; \(\frac{\mathrm{b}_1}{\mathrm{b}_2}\) = \(\frac{2}{4}\) = \(\frac{1}{2}\); \(\frac{\mathrm{c}_1}{\mathrm{c}_2}\) = \(\frac{6}{18}\) = \(\frac{1}{3}\)
∴ \(\frac{\mathrm{a}_1}{\mathrm{a}_2}\) = \(\frac{\mathrm{b}_1}{\mathrm{b}_2}\)
⇒ The lines are parallel and hence no solution.
3x + 2y = 6 ⇒ 2y = 6 – 3x
⇒ y = \(\frac{6-3 x}{2}\)

6x + 4y = 18 ⇒ 4y = 18 – 6x
⇒ y = \(\frac{18-6 \mathrm{x}}{4}\)


Scale : X – axis – 1 unit = 1 cm
Y – axis – 1 unit = 1 cm

Do Thiš

Question 1.
Solve each pair of equations by using the substitution method. (Page No. 88)
i) 3x – 5y = -1
x – y = -1
Solution:
Given: 3x – 5y = -1 —– (1)
x – y = -1 —– (2)
From Equation (2),
x – y = -1
x = y – 1
Substituting x = y – 1 in equation (1), we get
3(y – 1) – 5y = -1
⇒ 3y – 3 – 5y = -1
⇒ -2y = -1 + 3
⇒ 2y = -2
⇒ y = -1
Substituting y = – 1 in equation (1), we get
3x – 5(-1) = -1
3x + 5 = -1
3x = -6
x = -2
∴ The solution is (-2, -1)

ii) x + 2y = -1
2x – 3y = 12
Solution:
Given x + 2y = -1 —- (1)
2x – 3y = 12 —– (2)
From Equation (1), x + 2y = -1
⇒ x = -1 – 2y
Substituting x = -1 – 2y in equation (2), we get
2(-1 – 2y) – 3y = 12
– 2 – 4y – 3y = 12
-2 – 7y = 12
7y = -2 – 12
∴ y = \(\frac{-14}{7}\) = -2
Substituting y = -2 in equation (1), we get
x + 2(-2) = -1
x = -1 + 4
∴ x = 3
∴ The solution is (3, -2).

iii) 2x + 3y = 9
3x + 4y = 5
Solution:
Given : 2x + 3y = 9 —– (1)
3x + 4y = 5 —– (2)
From equation (1),
2x = 9 – 3y
⇒ x = \(\frac{9-3 y}{2}\)
Substituting x = \(\frac{9-3 y}{2}\) in equation (2), we get
3(\(\frac{9-3 y}{2}\)) + 4y = 5
⇒ \(\frac{27-9 y+2 \times 4 y}{2}\) = 5
⇒ 27 – 9y + 8y = 5 × 2
⇒ -y = 10 – 27
∴ y = 17
Substituting y = 17 in equation (1), we get
2x + 3 (+ 17) = 9
⇒ 2x = 9 – 51 ⇒ 2x = -42
⇒ x = -21
∴ The solution is (-21, 17).

iv) x + \(\frac{6}{y}\) = 6
3x – \(\frac{8}{y}\) = 5
Solution:
Given
x + \(\frac{6}{y}\) = 6 —- (1)
3x – \(\frac{8}{y}\) = 5 — (2)
From equation (1), x = 6 – \(\frac{6}{y}\)
Substituting x = 6 – \(\frac{6}{y}\) in equation (2), we get

Substituting y = 2 in equation (1), we get
x + \(\frac{6}{2}\) = 6
⇒ x + 3 = 6
∴ x = 3
∴ The solution is (3, 2).

v) 0.2x + 0.3y = 1.3
0.4x + 0.5y = 2.3
Solution:
Given:
0.2x + 0.3y = 1.3 ⇒ 2x + 3y = 13 —- (1)
0.4x + 0.5y = 2.3 ⇒ 4x + 5y = 23 —- (2)
From equation (1),
2x = 13 – 3y = x = \(\frac{13-3 y}{2}\)
Substituting x = \(\frac{13-3 y}{2}\) in equation (2), we get
4(\(\frac{13-3 y}{2}\)) + 5y = 23
⇒ 26 – 6y + 5y = 23
⇒ -y + 26 = 23
⇒ y = 26 – 23 = 3
Substituting y = 3 in equation (1). we get
⇒ 2x + 3(3) = 13
⇒ 2x + 9 = 13
⇒ 2x = 13 – 9
⇒ 2x = 4
⇒ x = 2
∴ The solution is (2, 3).

vi) \(\sqrt{2}\)x + \(\sqrt{3}\)y = 0
\(\sqrt{3}\)x – \(\sqrt{8}\)y = 0
Solution:
Given:
\(\sqrt{2}\)x + \(\sqrt{3}\)y = 0 —–(1)
\(\sqrt{3}\)x – \(\sqrt{8}\)y = 0 —— (2)

∴ The solution is x = 0, y = 0.
Note: a₁x + b₁y + c = 0
a₂x + b₂y + c₂ = 0
then, x = 0, y = 0 is a solution.

Question 2.
Solve each of the following pairs of equations by the elimination method. (Page No. 89)
i) 8x + 5y = 9
3x + 2y = 4
Solution:
Given : 8x + 5y = 9 —- (1)
3x + 2y = 4 —- (2)

Substituting y = 5 in equation (1), we get
8x + 5 × 5 = 9
⇒ 8x = 9 – 25
x = \(\frac{-16}{8}\)
∴ The solution is (-2, 5).

ii) 2x + 3y = 8
4x + 6y = 7
Solution:
Given: 2x + 3y = 8 —- (1)
4x + 6y = 7 —- (2)

The lines are parallel.
∴ The pair of lines has no solution.

iii) 3x + 4y = 25
5x – 6y = -9
Solution:
Given: 3x + 4y = 25 — (1)
5x – 6y = -9 — (2)

Substituting y = 4 in equation (1), we get
3x + 4 × 4 = 25
3x = 25 – 16
∴ x = \(\frac{9}{3}\) = 3
∴ (3, 4) is the solution for given pair of lines.

Question 3.
In a competitive exam, 3 marks are to be awarded for every correct answer and for every wrong answer, 1 mark will be deducted. Madhu scored 40 marks in this exam. Had 4 marks been awarded for each correct answer and 2 marks deducted for each incorrect answer, Madhu would have scored 50 marks. How many questions were there in the test ? (Madhu attempted all the questions). Use the elimination method. (Page No. 91)
Solution:
The equations formed are
3x – y = 40 — (1)
4x – 2y = 50 — (2)

Substituting y = 5 in equation (1), we get
3x – 5 = 40
3x = 40 + 5
x = \(\frac{45}{3}\) = 15
Total number of questions = Number of correct questions + Number of wrong answers
= x + y
= 15 + 5 = 20

Question 4.
Mary told her daughter, Seven years ago, I was seven times as old as you were then. Also, three years from now, I shall be three times as old as you will be. Find the present age of Mary and her daughter. Solve by the substitution method.
(Page No. 92)
Solution:
The Equation formed are
x – 7y + 42 = 0 —- (1)
x – 3y – 6 = 0 — (2)
From (1), x = -42 + 7y
Substituting in equation (2), we get
x = -42 + 7y
⇒ -42 + 7y – 3y – 6 = 0
⇒ 4y – 48 = 0
⇒ y = \(\frac{48}{4}\) = 12
Substituting y = 12 in equation (2), we get
x – 3 × 12 – 6 = 0
x – 36 – 6 = 0
∴ x = 42

Students can practice 10th Class Maths Study Material Telangana Chapter 11 Trigonometry InText Questions to get the best methods of solving problems.

TS 10th Class Maths Solutions Chapter 11 Trigonometry InText Questions

Do This

Identify “Hypotenuse”, “Opposite side” and “Adjacent side” for the given angles in the given triangles. (AS₃) (Page No. 271)

Question 1.
For angle R
Solution:

In the ∆PQR
Opposite side = PQ
Adjacent side = QR
Hypotenuse side = PR

Question 2.
i) For angle X
ii) For angle Y

Solution:
i) In the ∆XYZ
For angle X
Opposite side = YZ
Adjacent side = XZ
Hypotenuse = XY

ii) For angle Y
Opposite side = XZ
Adjacent side = YZ
Hypotenuse = XY

Question 3.
Find (i) sin C (ii) cos C (iii) tan C in the given triangle.
Solution:
By Pythagoras theorem
AC² = AB² + BC²
(13)² = AB² + (5)²
AB² = 169 – 25
AB² = 144
∴ AB = \(\sqrt{144}\) = 12

Question 4.
In a triangle XYZ, ZY is right angle. XZ = 17 cm and YZ = 15 cm, then find
(i) sin X (ii) cos Z (iii) tan X. (AS₁) (Page No. 274)
Solution:
Given ∆XYZ, ∠Y is right angle.
By Pythagoras theorem
XZ² = YZ² + XY²
17² = 15² + XY²
XY² = 17² – 15² = 289 – 225 = 64
XY = \(\sqrt{64}\) = 8

Question 5.
In a triangle PQR with right angle at Q, the value of ∠P is x, PQ = 7 cm and QR = 24 cm then find sin x and cos x. (AS₁) (Page No. 274)
Solution:
Given right angled triangle is PQR with right angle at Q. The value of ∠P is x.
By Pythagoras theorem
PR² = PQ² + QR²
= 7² + 24²

PR² = 49 + 576
PR² = 625
PR = \(\sqrt{625}\) = 25
sin x = \(\frac{\mathrm{QR}}{\mathrm{PR}}\) = \(\frac{24}{25}\)
cos x = \(\frac{\mathrm{PQ}}{\mathrm{PR}}\) = \(\frac{7}{25}\)

Try This

Question 1.
Write lengths of “Hypotenuse”, “Opposite side” and “Adjacent side” for the given angles in the given triangles. (AS₁) (Page No. 271)
i) For angle C
ii) For angle A
Solution:

In ∆ ABC, ∠B = 90°
∴ AC² = AB² + BC² (By Pythagoras theorem)
⇒ AB² = AC² – BC²
Substitute BC = 4 cm and AC = 5 cm in eq. (1), we get
AB² = 5² – 4² = 25 – 16 = 9
∴ AB = \(\sqrt{9}\) = 3
i) For angle C :
Length of hypotenuse = AC = 5 cm
Length of opposite side = AB = 3 cm
Length of adjacent side = BC = 4 cm

ii) For angle A :
Length of hypotenuse AC = 5 cm
Length of opposite side = BC = 4 cm
Length of adjacent side = AB = 3 cm

Question 2.
In a right angle triangle ABC, right angle is at C, BC + CA = 23 cm and BC – CA = 7 cm, then find sin A and tan B. (AS₁) (Page No. 274)
Solution:
∆ABC, ∠C = 90°
Given that,

⇒ BC = \(\frac{30}{2}\) = 15
BC + CA = 23
15 + CA = 23
CA = 23 – 15 = 8
AB² = BC² + CA² (By Pythagoras theorem)
= 15² + 8² = 225 + 64 = 289
∴ AB = \(\sqrt{289}\) = 17
sin A = \(\frac{\mathrm{BC}}{\mathrm{AB}}\) = \(\frac{15}{17}\) ; tan B = \(\frac{\mathrm{AC}}{\mathrm{BC}}\) = \(\frac{8}{15}\)

Question 3.
What will be the ratio of sides for sec A and cot A ? (AS₃) (Page No. 275)
Solution:
sec A = \(\frac{1}{\cos A}\) = \(\frac{\text { Hypotenuse } }{\text { Side opposite to angle A }}\)
cot A = \(\frac{1}{\tan A}\) = \(\frac{\text { Side adjacent to angle A} }{\text { Side opposite to angle A }}\)

Think – Discuss

Question 1.
Discuss between your friends that
i) sin x = \(\frac{4}{3}\) does exist for some value of angle x ?
ii) The value of sin A and cos A is always less than 1. Why ?
iii) tan A is product of tan and A.   (AS₂) (Page No. 274)
Solution:
i) The value of sin0 always lies between 0 and 1. Here, sin x = \(\frac{4}{3}\) which is greater than 1. So, it does not exist.

ii) Draw a circle of radius 1 unit with centre at the origin. Let P(a, b) be any point on the circle with angle AOP = θ
sin θ = \(\frac{\mathrm{AP}}{\mathrm{OP}}=\frac{\mathrm{b}}{1}\) = y – co-ordinate
cos θ = \(\frac{\mathrm{OA}}{\mathrm{OP}}=\frac{\mathrm{a}}{1}\) = x – co-ordinate
One complete rotation of point ‘P’ in a circular an angle 360° at the centre.

Quarter rotation substends an angle of AOB equals to 90°.
Half rotation substends an angle of AOC equal to 180°.
Three quarter substends an angle of AOD equals to 270°.
The co-ordinates of the points A, B, C and D respectively (1, 0), (0, 1) (-1, 0) and (0, -1).
According to the co-ordinates
cos 0° = 1 sin 0° = 0
cos 90° = 0 sin 90° = 1
cos 180° = – 1 sin 180° = 0
cos 270° = 0 sin 270° = -1
cos 360° = 1 sin 360° = 0

Hence, the value of “sine” and “cosine” is always less than 1.

iii) The symbol tan A is used as an abbreviation for “the tan of the angle A”, tan A is not the product of “tan” and A “tan” separated from “A” has no meaning.

Question 2.
Is \(\frac{\sin \mathrm{A}}{\cos \mathrm{A}}\) equal to tan A ? (Page No. 275)
Solution:

Question 3.
Is \(\frac{\cos \mathrm{A}}{\sin \mathrm{A}}\) equal to cot A ? (Page No. 275)
Solution:

Do This

Question 1.
Find cosec 60°, sec 60° and cot 60°. (AS₁) (Page No. 279)
Solution:
Consider an equilateral triangle ABC. Since each angle is 60° in an equilateral triangle, we have ∠A = ∠B = ∠C = 60° and the sides of equilateral triangle is AB = BC = CA = 2a units

Draw the perpendicular line AD from vertex A to BC as shown in the given figure.

Perpendicular AD acts as “angle bisector of angle A” and “bisector of the side BC” in the equilateral triangle ABC.
Therefore, ∠BAD = ∠CAD = 30°.
Since point D divides the side BC into equal halves.
BD = \(\frac{1}{2}\) BC = \(\frac{2 \mathrm{a}}{2}\) = a units
Consider right angle triangle ABD in the above given figure.
We have AB = 2a, and BD = a
Then AD² = AB² – BD² (By Pythagoras theorem)
= (2a)² – (a)² = 3a²
Therefore, AD = a\(\sqrt{3}\)
From, definitions of trigonometric ratios.
sin 60° = \(\frac{\mathrm{AD}}{\mathrm{AB}}=\frac{\mathrm{a} \sqrt{3}}{2 \mathrm{a}}=\frac{\sqrt{3}}{2}\)
cos 60° = \(\frac{\mathrm{BD}}{\mathrm{AB}}=\frac{\mathrm{a}}{2 \mathrm{a}}=\frac{1}{2}\)
So, similarly tan 60° = \(\sqrt{3}\)

Try This

Question 1.
Find sin 30°, cos 30°, tan 30°, cosec 30°, sec 30° and cot 30° by using the ratio concepts. (AS₁) (Page No. 279)
Solution:
Let ABC be an equilateral triangle.
∴ ∠A = ∠B = ∠C = 60°
Let AB BC = CA ‘a’ units
Draw AD ⊥ BC, AD bisects BC.
∴ BD = a/2
In ∆ABD, ∠ADB = 90°; ∠B = 60°
∴ ∠BAD = 180° – (90° + 60°)
= 180° – 150° = 30°
By Pythagoras theorem,
AB² = AD² + BD²
⇒ AD² = AB² – BD²

Question 2.
Find the values for tan 90°, cosec 90°, sec 90° and cot 90°. (AS₁) (Page No. 281)
Solution:
Let us see what happens when angle made by AC with ray AB increases. When angle A is increased, height of point C increases and the foot of the perpendicular shifts from B to X and then to Y and so on. So, when the angle becomes 90°, base (adjacent side of the angle) would become zero; the height of C from AB ray increases and it would be equal to AC.
AB = 0 and BC = AC

Think – Discuss

Question 1.
Discuss between your friend about the following conditions :
What can you say about cosec 0° = \(\frac{1}{\sin 0}\) ? Is it not defined ? Why ? (AS₂) (Page No. 280)
Solution:
sin 0° = 0
cosec 0° = \(\frac{1}{\sin 0}\) = \(\frac{1}{0}\) = not defined
Reason : Division by ‘0’ is not allowed, hence \(\frac{1}{0}\) is indeterminate.

Question 2.
Is it defined cot 0° = \(\frac{1}{\tan 0}\) ? Why ? (Page No. 281)
Solution:
tan 0° = 0
cot 0° = \(\frac{1}{\tan 0}\) = \(\frac{1}{0}\) undefined
Reason : Division by ‘0’ is not allowed, hence \(\frac{1}{\tan 0}\) is indeterminate.

Question 3.
sec 0° = 1. Why ? (Page No. 281)
Solution:
sec 0° = \(\frac{1}{\cos 0^{\circ}}\) [∵ cos 0° = 1]
= \(\frac{1}{1}\) = 1

Question 4.
What can you say about the values of sin A and cos A, as the value of angle A increases from 0° to 90° ? (AS₃) (Page No. 282)
i) If A > B, then sin A > sin B. Is it true ?
ii) If A > B, then cos A > cos B. Is it true ? Discuss.
Solution:
i) Given statement
“If A > B, then sin A > sin B”.
Yes, this statement is true.
Because, it is clear from the table below that the sin A increases as A increases.

ii) Given statement
“If A > B, then cos A > cos B”.
No, this statement is not true.
Because it is clear from the table below that cos A decreases as A increases.

Think – Discuss

Question 1.
For which value of acute angle
i) \(\frac{\cos \theta}{1-\sin \theta}+\frac{\cos \theta}{1+\sin \theta}\) = 4 is true ? For which value of 0° ≤ θ ≤ 90°, above equa-tion is not defined ? (AS₁, AS₂) (Page No. 285)
Solution:
Given \(\frac{\cos \theta}{1-\sin \theta}+\frac{\cos \theta}{1+\sin \theta}\) = 4

cos θ = \(\frac{1}{2}\)
cos θ = cos 60°
∴ θ = 60°
Given statement is true for the acute angle i.e., θ = 60°

Question 2.
Check and discuss the above relations AB
[sin (90° – x) = \(\frac{\mathrm{A B}}{\mathrm{A C}}\) = cos x and BC
cos (90° – x) = \(\frac{\mathrm{B C}}{\mathrm{A C}}\) = sin x,
tan (90° – x) = \(\frac{\mathrm{A B}}{\mathrm{A C}}\) = cot x and
cot (90° – x) = \(\frac{\mathrm{B C}}{\mathrm{A B}}\) = tan x,
cosec (90° – x) = \(\frac{\mathrm{A C}}{\mathrm{A B}}\) = sec x and
sec (90° – x) = \(\frac{\mathrm{A C}}{\mathrm{B C}}\) = cosec x] in the case of angles between 0° and 90°, whether they hold for these angles or not ? So,
(i) sin (90° – A) = cos A
(ii) cos (90° – A) = sin A
(iii) tan (90° – A) = cot A and
(iv) cot (90° – A) = tan A
(vi) cosec (90° – A) = cosec A
(v) sec (90° – A) = cosec A (AS₂) (Page No. 286)
Solution:
Let A = 30°
i) sin (90° – A ) = cos A
⇒ sin (90° – 30°) = cos 30°
⇒ sin 60° = cos 30° = \(\frac{\sqrt{3}}{2}\)

ii) cos (90° – A ) = sin A
⇒ cos (90° – 30°) = sin 30°
⇒ cos 60° – sin 30° = \(\frac{1}{2}\)

iii) tan (90° – A ) = cot A
⇒ tan (90° – 30°) = cot 30°
⇒ tan 60° = cot 30° = \(\sqrt{3}\)

iv) cot (90° – A ) = tan A
⇒ cot (90° – 30°) = tan 30°
⇒ cot 60° = tan 30° = \(\frac{1}{\sqrt{3}}\)

v) sec (90° – A ) = cosec A
⇒ sec (90° – 30°) = cosec 30°
⇒ sec 60° = cosec 30° = 2

vi) cosec (90° – A ) = sec A
⇒ cosec (90° – 30°) = sec 30°
⇒ cosec 60° = sec 30° = \(\frac{2}{\sqrt{3}}\)
So, the above relations hold for all the angles between 0° and 90°.

Do This

i) If sin C = \(\frac{15}{17}\), then find cos C. (AS₁) (Page No. 290)
Solution:
Given sin C = \(\frac{15}{17}\)
cos C = \(\sqrt{1-\sin ^2 \mathrm{C}}\) (from identity – 1)
= \(\sqrt{1-\left(\frac{15}{17}\right)^2}\) = \(\sqrt{\frac{289-225}{289}}\) = \(\sqrt{\frac{64}{289}}\) = \(\frac{8}{17}\)

ii) If tan x = \(\frac{5}{12}\), then find sec x. (AS₁) (Page No. 290)
Solution:
Given tan x = \(\frac{5}{12}\)
We know that sec²x – tan²x = 1
sec²x = 1 + tan²x

iii) If cosec θ = \(\frac{25}{7}\), then find cot θ. (AS₁) (Page No. 290)
Solution:
Given cosec θ = \(\frac{25}{7}\)
We know that cosec²θ – cot²θ = 1
cot²θ = cosec²θ – 1

Try This

Question 1.
Evaluate the following and justify your answer. (AS₄) (Page No. 290)
i) \(\frac{\sin ^2 15+\sin ^2 75}{\cos ^2 36+\cos ^2 54}\)
ii) sin 5° cos 85° + cos 5° sin 85°
iii) sec 16° cosec 74° – cot 74° tan 16°
Solution:
i) \(\frac{\sin ^2 15+\sin ^2 75}{\cos ^2 36+\cos ^2 54}\)
We can write sin 15°= sin (90° – 75°)
= cos 75°
∴ sin² 15° = cos² 75°
Similarly, cos 36°= cos (90° – 54°) = sin 54°
∴ cos² 36° = sin² 54°
sin² 15° + sin² 75°= cos² 75° + sin² 75°
(∵ sin² 15° = cos² 75°)
= 1 ———– (1) (∵ cos²θ + sin²θ = 1)
(Here θ = 75°)
cos² 36° + cos² 54°= sin² 54° + cos² 54°
(∵ cos² 36° = sin² 54°)
= 1 ———– (2) (∵ sin²θ + cos²θ = 1)
(Here θ = 54°)
From (1) & (2), we get
\(\frac{\sin ^2 15+\sin ^2 75}{\cos ^2 36+\cos ^2 54}\) = \(\frac{1}{1}\) = 1

ii) sin 5° cos 85° + cos 5° sin 85° …………….. (1)
sin 5°= sin (90° – 85°) = cos 85°
cos 5° = cos (90° – 85°) = sin 85°
Substitute these values of sin 5° and cos 5° in (1).
We get
sin 5° cos 85° + cos 5° sin 85°
= cos 85°. cos 85° + sin 85°. sin 85°
= cos² 85° + sin² 85°
= 1 (∵ cos² θ + sin² θ = 1, Here θ = 85°)

iii) sec 16° cosec 74° – cot 74° tan 16° …………….. (1)
cosec 74° = cosec (90° – 16°) = sec 16°
[(∵ cosec (90° – θ) = sec θ) and cot(90° – θ) = tan θ]
cot 74° = cot (90° – 16°) = tan 16°
Substitute the equivalents of cosec 74° and cot 74° in (1), we get
sec 16° . cosec 74° – cot 74° . tan 16°
= sec 16° . sec 16° – tari 16° . tan 16°
= sec² 16° – tan² 16° = 1
(∵ sec² θ – tan²θ = 1) Here θ = 16°

Think – Discuss

Question 1.
Are these identities true for 0° ≤ A ≤ 90° ? If not, for which values of A they are true ?
i) sec² A – tan² A = 1
ii) cosec² A – cot² A = 1 (AS₂) (Page No. 290)
Solution:
i) Given identity is sec² A – tan² A = 1
Let A = 0°
LHS = sec² 0° – tan² 0°
= 1 – 0 = 1 = R.H.S.
Let A =90°
tan A and sec A are not defined.
So it is true.
∴ For all given values of ‘A’ such that 0° ≤ A ≤ 90° this trigonometric identity is true.

ii) Given identity is cosec² A – cot² A = 1
Let A = 0°
cosec A and cot A are not defined for A = 0°.
Therefore identity is true for A = 0°
Let A = 90°
cosec A = cosec 90° = 1
cot A = cot 90° = 0
∴ L.H.S. = 1² – 0² = 1 – 0 = 1 = R.H.S.
∴ This identity is true for all values of A, such that 0° ≤ A ≤ 90°

Students can practice 10th Class Maths Study Material Telangana Chapter 11 Trigonometry Optional Exercise to get the best methods of solving problems.

TS 10th Class Maths Solutions Chapter 11 Trigonometry Optional Exercise

Question 1.
Prove that \(\frac{\cot \theta-\cos \theta}{\cot \theta+\cos \theta}\) = \(\frac{{cosec} \theta-1}{{cosec} \theta+1}\)
Solution:

(Dividing the numerator and denominator by ‘sin θ’)

Question 2.
Prove that \(\frac{\sin \theta-\cos \theta+1}{\sin \theta+\cos \theta-1}\) = \(\frac{1}{\sec \theta-\tan \theta}\) using the identify sec²θ = 1 + tan²θ.
Solution:

Question 3.
Prove that (cosec A – sin A) (sec A – cos A) = \(\frac{1}{\tan A+\cot A}\)
Solution:
L.H.S. = (cosec A – sin A) (sec A – cos A)
= (\(\frac{1}{\sin \mathrm{A}}\) – sin A) (\(\frac{1}{\cos \mathrm{A}}\) – cos A)
(∵ sin²A + cos²A = 1 ⇒ 1 – sin²A = cos²A and 1 – cos²A = sin²A)

∴ L.H.S. = R.H.S
Hence (cosec A – sin A) (sec A – cos A)
= \(\frac{1}{\tan A+\cot A}\)

Question 4.
Prove that \(\frac{1+\sec \mathrm{A}}{\sec \mathrm{A}}\) = \(\frac{\sin ^2 A}{1-\cos A}\)
Solution:

(∵ sin²A + cos²A = 1 ⇒ sin²A = 1 – cos²A)

Question 5.
Show that \(\left(\frac{1+\tan ^2 \mathrm{~A}}{1+\cot ^2 \mathrm{~A}}\right)\) = \(\left(\frac{1+\tan A}{1-\cot A}\right)^2\) = tan²A
Solution:

Question 6.
Prove that \(\frac{(\sec A-1)}{(\sec A+1)}\) = \(\frac{(1-\cos \mathrm{A})}{(1+\cos \mathrm{A})}\)
Solution:

Students can practice 10th Class Maths Study Material Telangana Chapter 11 Trigonometry Ex 11.4 to get the best methods of solving problems.

TS 10th Class Maths Solutions Chapter 11 Trigonometry Exercise 11.4

Question 1.
Evaluate the following :
i) (1 + tan θ + sec θ) (1 + cot θ – cosec θ)
ii) (sin θ + cos θ)² + (sin θ – cos θ)²
iii) (sec² θ – 1) (cosec² θ – 1) (AS₁)
Solution:
i) (1 + tan θ + sec θ) (1 + cot θ – cosec θ)
= (1 + tan θ + sec θ)

ii) (sin θ + cos θ)² + (sin θ – cos θ)²
(sin θ + cos θ)² = sin² θ + cos² θ + 2 sin θ cos θ …………. (1)
(sin θ – cos θ)² = sin² θ + cos² θ – 2 sin θ cos θ ……………. (2)
Adding (1) & (2)
sin² θ + cos² θ + 2 sin θ (cos θ) + sin² θ + cos² θ – 2 sin θ (cos θ)
= 2 (sin² θ + cos² θ)
= 2(1) = 2

iii) (sec² θ – 1) (cosec² θ – 1)
sec² θ – 1 = tan² θ …………… (1)
[∵ 1 + tan² θ = sec² θ ⇒ tan² θ = sec² θ – 1]
cosec² θ – 1 = cot² θ …………… (2)
[∵ 1 + cot² θ = cosec² θ ⇒ cot² θ = cosec² θ – 1]
⇒ tan² θ cot² θ
⇒ (tan θ cot θ)² [∵ (tan θ) (cot θ) = 1]
⇒ (1)² = 1

Question 2.
Show that   (A.P. June ’15)
(cosec θ – cot θ)² = \(\frac{1-\cos \theta}{1+\cos \theta}\) (AS₂)
Solution:
(cosec θ – cot θ)² = \(\left(\frac{1}{\sin \theta}-\frac{\cos \theta}{\sin \theta}\right)^2\)

Question 3.
Show that \(\sqrt{\frac{1+\sin A}{1-\sin A}}\) = sec A + tan A. (AS₂)
Solution:

Question 4.
Show that \(\frac{1-\tan ^2 A}{\cot ^2 A-1}\) = tan² A. (AS₂)
Solution:

Question 5.
Show that \(\frac{1}{\cos \theta}\) – cos θ = tan θ . sin θ (AS₂)
Solution:
L.H.S. = \(\frac{1}{\cos \theta}\) = cos θ = \(\frac{1-\cos ^2 \theta}{\cos \theta}\)
= \(\frac{\sin ^2 \theta}{\cos \theta}\)
[∵ sin²θ + cos²θ = 1 ⇒ sin²θ = 1 – cos²θ]
= \(\frac{\sin \theta}{\cos \theta}\) . sin θ
= tan θ . sin θ (∵ tan θ = \(\frac{\sin \theta}{\cos \theta}\))

Question 6.
Simplify: sec A (1 – sin A). (sec A + tan A). (AS₃)
Solution:
sec A (1 – sin A) = (sec A – sec A . sin A)
= (sec A – \(\frac{\sin A}{\cos A}\))
= (sec A – tan A)
∴ sec A (1 – sin A) = (sec A + tan A)
= (sec A + tan A) (sec A – tan A)
= sec² A – tan² A
= 1
[∵ 1 + tan² A = sec² A ⇒ sec² A – tan² A = 1]

Question 7.
Prove that (sin A + cosec A)² + (cos A + sec A)² = 7 + tan² A + cot² A) (AS₁, AS₂)
Solution:
L.H.S. = (sin A + cosec A)² + (cos A + sec A)²
= sin² A + cosec² A + 2 sin A. cosec A + cos² A + sec² A + 2 cos A . sec A
= (sin² A + cos² A) + cosec² A+ 2 + sec² A + 2
= 1 + cosec² A + 2 + sec² A + 2
[∵ sin² A + cos² A = 1]
= 5 + cosec² A + sec² A
= 5 + 1 + cot² A + 1 + tan² A
(∵ 1 + cot² A = cosec² A and 1 + tan² A = sec² A)
= 7 + tan² A + cot² A

Question 8.
Simplify : (T.S. Mar.’15) (AS₁, AS₃)
(1 – cos θ) (1 + cos θ) (1 + cot² θ)
Solution:
(1 – cos θ) (1 + cos θ) (1 + cot² θ)
= (1 – cos² θ) (1 + cot² θ)
[∵ (a – b) (a + b) = a² – b²]
= sin² θ (1 + cot² θ)
(∵ sin² θ + cos² θ = 1 ⇒ sin² θ = 1- cos² θ)
= sin² θ + sin² θ . cot² θ
= sin² θ + sin² θ . \(\frac{\cos ^2 \theta}{\sin ^2 \theta}\)
(∵ cot θ = \(\frac{\cos \theta}{\sin \theta}\) cot²θ = \(\frac{\cos ^2 \theta}{\sin ^2 \theta}\))
= sin² θ + cos² θ
= 1

Question 9.
If sec θ + tan θ = p, then what is the value of sec θ – tan θ ? (AS₁)
Solution:
We know that 1 + tan² θ = sec2 θ
⇒ sec² θ – tan² θ = 1
⇒ (sec θ + tan θ) (sec θ – tan θ) = 1
[∵ a² – b² = (a + b) (a – b)]
⇒ p(sec θ – tan θ) = 1
⇒ sec θ – tan θ = \(\frac{1}{\mathrm{p}}\).
∴ The value of sec θ – tan θ = \(\frac{1}{\mathrm{p}}\)

Question 10.
If cosec θ + cot θ = k, then prove that cos θ = \(\frac{\mathrm{k}^2-1}{\mathrm{k}^2+1}\). (A.P. Mar.’16) (AS₁)
Solution:
Given that cosec θ – cot θ = k

Students can practice 10th Class Maths Study Material Telangana Chapter 11 Trigonometry Ex 11.3 to get the best methods of solving problems.

TS 10th Class Maths Solutions Chapter 11 Trigonometry Exercise 11.3

Question 1.
Evaluate. (AS₄)

i) \(\frac{\tan 36}{\cot 54}\)
Solution:
Given that \(\frac{\tan 36}{\cot 54}\)
= \(\frac{\tan 36}{\cot \left(90^{\circ}-36^{\circ}\right)}\) = \(\frac{\tan 36^{\circ}}{\tan 36^{\circ}}\) = 1

ii) cos 12° – sin 78°
Solution:
Given that cos 12° – sin 78°
= cos 12° – sin (90° – 12°)
= cos 12° – cos 12°
= 0 [∵ sin (90° – θ) = cos θ]

iii) cosec 31° – sec 59°
Solution:
Given that cosec 31° – sec 59°
= cosec 31° – sec (90° – 31°) [∵ sec (90°-31°) = cosec 31°]
= cosec 31° – cosec 31°
= 0

iv) sin 15° sec 75°
Solution:
Given that sin 15° sec 75°
= sin 15° sec (90° – 15°)
= sin 15° cosec 15° [∵ sec (90° – θ) = cosec θ]
1 sin 15°
= 1

v) tan 26° tan 64°
Solution:
Given that tan 26° tan 64°
= tan 26°. tan (90° – 26°)
= tan 26° . cot 26° [∵ tan (90° – θ) = cot θ]
= tan 26° . \(\frac{1}{\tan 26^{\circ}}\)
= 1

Question 2.
Show that (AS₂)
i) tan 48° tan 16° tan 42° tan 74° = 1
Solution:
L.H.S. = tan 48° tan 16° tan 42° tan 74°
= tan 48°. tan 16°. tan (90° – 48°) . tan (90° – 16°)
= tan 48°. tan 16°. cot 48°. cot 16° [∵ tan (90 – θ) = cot θ]
= tan 48° tan 16° \(\frac{1}{\tan 48^{\circ}}\) \(\frac{1}{\tan 16^{\circ}}\) [∵ cot θ = \(\frac{1}{\tan \theta}\)]
= 1 = R.H.S.

ii) cos 36° cos 54° – sin 36° sin 54° = 0
Solution:
L.H.S. = cos 36° cos 54° – sin 36° sin 54°
= cos (90° – 54°). cos (90° – 36°) – sin 36°. sin 54°
= sin 54°. sin 36° – sin 36°. sin 54° [∵ cos (90 – θ) = sin θ]
= 0 = R.H.S.
∴ L.H.S. = R.H.S.

Question 3.
If tan 2A = cot (A – 18°), where 2A is an acute angle. Find the value of A. (AS₁, AS₄)
Solution:
Given that, tan 2A = cot (A – 18°)
⇒ cot (90° – 2A) = cot (A – 18°) [∵ tan θ = cot (90 – θ)]
⇒ 90° – 2A = A – 18°
⇒ 108° = 3A
∴ A = \(\frac{108^{\circ}}{3}\) = 36°
Hence, the value of ‘A’ is 36°

Question 4.
If tan A = cot B where A and B are acute angles, prove that A + B = 90°. (AS₂)
Solution:
Given that tan A = cot B
⇒ cot (90° – A) = cot B
[∵ tan θ = cot (90 – θ)]
⇒ 90° – A = B
∴ A + B = 90°

Question 5.
If A, B and C are interior angles of a triangle ABC, then show that tan \(\left(\frac{\mathrm{A}+\mathrm{B}}{2}\right)\) = cot \(\left(\frac{\mathrm{C}}{2}\right)\)
Solution:
Given A, B and C are interior angles of right angle triangle ABC then
A + B + C = 180°
On dividing the above equation by ‘2’ on both sides, we get
\(\frac{\mathrm{A}+\mathrm{B}}{2}\) + \(\frac{\mathrm{C}}{2}\) = \(\frac{180^{\circ}}{2}\) = 90°
\(\frac{\mathrm{A}+\mathrm{B}}{2}\) = 90° – \(\frac{\mathrm{C}}{2}\)
On taking ‘tan’ ratio on both sides.
tan \(\left(\frac{\mathrm{A}+\mathrm{B}}{2}\right)\) = tan \(\left(90^{\circ}-\frac{C}{2}\right)\)
tan \(\left(\frac{\mathrm{A}+\mathrm{B}}{2}\right)\) = cot \(\frac{\mathrm{C}}{2}\)
Hence proved.

Question 6.
Express sin 75° + cos 65° in terms of trigonometric ratios of angles between 0° and 45°. (AS₃)
Solution:
We have sin 75° + cos 65°
= sin (90° – 15°) + cos (90° – 25°)
= cos 15° + sin 25°
[∵ sin (90 – θ) = cos θ and
cos (90 – θ) = sin θ]

Students can practice TS 10th Class Maths Solutions Chapter 4 Pair of Linear Equations in Two Variables Optional Exercise to get the best methods of solving problems.

TS 10th Class Maths Solutions Chapter 4 Pair of Linear Equations in Two Variables Optional Exercise

Question 1.
Solve the following equations :

i) \(\frac{2 x}{a}\) + \(\frac{\mathbf{y}}{\mathbf{b}}\) = 2
\(\frac{x}{a}\) – \(\frac{y}{b}\) = 4
Solution:
Given
\(\frac{2 x}{a}\) + \(\frac{\mathbf{y}}{\mathbf{b}}\) = 2 —- (1)
\(\frac{x}{a}\) – \(\frac{y}{b}\) = 4 — (2)
Adding eq.(1) & (2) \(\frac{3 x}{a}\) = 6 ⇒ x = \(\frac{6 a}{3}\) = 2a
Substituting x = 2a in the equation (1), we get
\(\frac{2}{a}\)(2a) + \(\frac{\mathrm{y}}{\mathrm{b}}\) = 2
⇒ 4 + \(\frac{\mathrm{y}}{\mathrm{b}}\) = 2
⇒ \(\frac{\mathrm{y}}{\mathrm{b}}\) = -2 ⇒ y = -2b
∴ The solution (x, y) = (7, 13)

ii) \(\frac{x+1}{2}\) + \(\frac{y-1}{3}\) = 8
\(\frac{x-1}{3}\) + \(\frac{y+1}{2}\) = 9
Solution:

Substituting y = 13 in equation (1), we get
3x + 2(13) = 47 ⇒ 3x = 47 – 26
⇒ 3x = 21 ⇒ x = \(\frac{21}{3}\) = 7
∴ The solution (x, y) = (7, 13)

iii) \(\frac{x}{7}\) + \(\frac{y}{3}\) = 5
\(\frac{x}{2}\) – \(\frac{y}{9}\) = 6
Solution:

Substituting y = 9 in equation (1) we get
3x + 7(9) = 105
⇒ 3x = 105 – 63
⇒ 3x = 42
⇒ x = \(\frac{42}{3}\) = 14
∴ The solution (x, y) = (14, 9)

iv) \(\sqrt{3}\)x – \(\sqrt{2}\)y = \(\sqrt{3}\)
\(\sqrt{5}\)x + \(\sqrt{3}\)y = \(\sqrt{3}\)
Solution:

v) \(\frac{a x}{\mathbf{b}}\) – \(\frac{b y}{a}\) = a + b
ax – by = 2ab
Solution:
Given \(\frac{a x}{\mathbf{b}}\) – \(\frac{b y}{a}\) = a + b —- (1)
ax – by = 2ab —- (2)

Substituting y = -a in equation (2), we get
ax – b(-a) = 2ab
⇒ ax + ab = 2ab
⇒ ax = 2ab – ab
∴ x = \(\frac{a b}{a}\) = b
∴ The solution (x, y) = (b, -a)

vi) 2^x + 3^y = 17
2^x+2 – 3^y+1 = 5
Solution:
Given, 2^x + 3^y = 17 and
2^x+2 – 3^y+1 = 5
Take 2^x = a and 3^y = b then the give equations reduce to
2^x + 3^y = 17 —- (1)
2^x.2² – 3^y. 3 = 5 ⇒ 4a – 3b = 5 —– (2)

Substituting b = 9 in equation (1), we get
a + 9 = 17 ⇒ a = 17 – 9 = 8
But a = 2^x = 8 and b = 3^y = 9
⇒ 2^x = 2³
⇒ x = 3

⇒ 3^y = 3²
⇒ y = 2
∴ The solution (x, y) is (3, 2).

Question 2.
Animals in an experiment are to be kept on a strict diet. Each animal is to receive among other things 20g of protein and 6g of fat. The laboratory technicians purchased two food mixes, A and B. Mix A has 10% protein and 6% fat. Mix B has 20% protein and 2% fat. How many grams of each mix should be used?
Solution:
Let x gms of mix A and y gms of mix B are to be mixed, then


∴ y = \(\frac{300}{5}\) = 60 gm
Substituting y = 60 in equation (1), we get
x + 2 × 60 = 200 ⇒ x + 120 = 200
⇒ 200 – 120 = 80 gm
∴ Quantity of mix. A = 80 gms.
Quantity of mix. B = 60 gms.

Students can practice 10th Class Maths Solutions Telangana Chapter 3 Polynomials InText Questions to get the best methods of solving problems.

TS 10th Class Maths Solutions Chapter 3 Polynomials InText Questions

Do this

Question 1.
State which of the following are polynomials and which are not ? Give reasons. (Page No. 48)
(i) 2x³
(ii) \(\frac{1}{x-1}\)
(iii) 4z² + \(\frac{1}{7}\)
(iv) m² – \(\sqrt{2}\) m + 2
(v) p^-2 + 1
Solution:
i) 2x³ is a polynomial
ii) \(\frac{1}{x-1}\) is not a polynomial.
iii) 4z² + \(\frac{1}{7}\) is a polynomial.
iv) m² – \(\sqrt{2}\)m + 2 is a polynomial.
v) p^-2 + 1 is not a polynomial.

Question 2.
p(x) = x² – 5x – 6, find the values of p(1), p(2), p(3), p(0), p(-1), p(-2), p(-3). (Page No. 49)
Solution:
p(x) = x² – 5x – 6
p(1) = (1)² – 5(1) – 6 = 1 – 5 – 6 = 1 – 11 = -10
p(2) = (2)² – 5(2) – 6 = 4 – 10 – 6
= 4 – 16 = – 12
p(3) = (3)² – 5(3) – 6 = 9 – 15 – 6
= 9 – 21 = -12
p(0) = (0)² – 5(0) – 6 = 0 – 0 – 6 = 0 – 6 = -6
p(-1) = (-1)² – 5(-1) – 6 = 1 + 5 – 6 = 6 – 6 = 0
p(-2) = (-2)² – 5(-2) – 6 = 4 + 10 – 6 = 8
p(-3) = (-3)² – 5(-3) – 6 = 9 + 15 – 6
= 24 – 6 = 18

Question 3.
p(m) = m² – 3m + 1, find the value of p(1) and p(-1). (Page No. 49)
Solution:
p(m) = m² – 3m + 1
p(1) = (1)² – 3(1) + 1 = 1 – 3 + 1 = 2 – 3 = -1
p(-1) = (-1)² – 3(-1) + 1 = 1 + 3 + 1 = 5

Question 4.
Let p(x) = x² – 4x + 3. Find the value of p(0), p(1), p(2), p(3) and obtain zeroes of the polynomial p(x).
(Page No. 50)
Solution:
p(x) = x² – 4x + 3
p(0) = (0)² – 4(0) + 3 = 0 – 0 + 3 = 3
p(1) = (1)² – 4(1) + 3 = 1 – 4 + 3 = 4 – 4 = 0
p(2) = (2)² – 4(2) + 3 = 4 – 8 + 3 = 7 – 8 = -1
p(3) = (3)² – 4(3) + 3 = 9 – 12 + 3 = 12 – 12 = 0
We see that p(1) = 0 & p(3) = 0
These points x = 1 & x = 3 are called zeroes of the polynomial p(x) = x² – 4x + 3

Question 5.
Check whether -3 and 3 are the zeroes of the polynomial x² – 9. (Page No. 50)
Solution:
p(x) = x² – 9 ⇒ p(-3) = (-3)² – 9 = 9 – 9 = 0
p(3) = (3)² – 9 = 9 – 9 = 0
p(-3) = 0 & p(3) = 0
∴ -3 and 3 are the zeroes of the polynomial p(x) = x² – 9

Try This

Question 1.
Write 3 different quadratic, cubic and 2 linear polynomials with different number of terms. (Page No. 48)
Solution:
3 different quadratic polynomials :

1. 3x² + 2x + 4
2. 5x² + x + 1
3. x² – 3x + 2

3 different cubic polynomials :

1. x³ + 2x² + x + 1
2. 3x³ + 2x + 1
3. x³ + 1

2 linear polynomials :

1. 4x + 3
2. 3x + 2

Question 2.
Write a quadratic polynomial and a cubic polynomial in variable x in the general form. (Page No. 49)
Solution:
General form of a (in variable x)

1. Quadratic polynomial → ax² + bx +c.
2. Cubic polynomial → ax³ + bx² + cx + d

Question 3.
Write a general polynomial q(z) of degree n with coefficients that are b₀,…,b_n. What are the conditions on b₀, ..,b_n ? (Page No. 49)
Solution:
General polynomial q(z) of degree ‘n’ is q(z) = b₀xⁿ + b₁x^n-1 + b₂x^n-2 + b₃x^n-3 +…. +b_n-1x + b_n

Conditions :

1. b₀, b₁, b₂, …….,b_n-1, b_n are real coefficients.
2. bg ≠ 0

Do This

Question 1.
Draw the graph of
(i) y = 2x + 5
(ii) y = 2x – 5
(iii) y = 2x and find the point of intersection on X-axis. Is the x- coordinate of these points also the zero of the polynomial ? (Page No. 52)

i) y = 2x +


The zero of the polynomial 2x + 5 is the x-coordinate of the point where the graph of y = 2x + 5 intersects the X-axis.
∴ Zero of the polynomial 2x + 5 is \(\frac{-5}{2}\).

ii) y = 2x – 5

The zero of the polynomial 2x – 5 is the x-coordinate of the point where the graph of y = 2x – 5 intersects the X-axis.
∴ Zero of the polynomial 2x – 5 is 5/2.

iii) y = 2x

The zero of the polynomial 2x is the x-coordinate of the point where the graph of y = 2x intersects the X-axis.
The graph y = 2x does not intersect X-axis.
∴ There is no zero of the polynomial for 2x.

Try This

Question 1.
Draw the graphs of
(i) y = x² – x – 6
(ii) y = 6 – x – x² and find zeroes in each case. What do you notice ? (A.P. Mar. 15) (Page No. 53)
Solution:
i) y = x² – x – 6

Zeroes of the quadratic polynomial x² – x – 6 are the x-co-ordinates of the points of X-axis.

-2 & 3 are zeroes of the quadratic polynomial and -2 & 3 are intersections points of X-axis.

ii) y = 6 – x – x²

Zero of the quadratic polynomial 6 – x – x² are the x-coordinates of the points of X-axis.
-3 & 2 are zeroes of the quadratic polynomial and -3 & 2 are intersection points of X-axis.

Observations : The graph cuts X-axis at two distinct points.
The parabola can open either upward or down-ward.

Question 2.
Write three quadratic polynomials that have 2 zeroes each. (Page No. 55)
Answer:

1. y = x² + 5x + 6
2. y = x² – 5x + 6
3. y = 2x² + x – 1

The above three polynomials have 2 zeroes each.

Question 3.
Write one quadratic polynomial that has one zero. (Page No. 55)
Answer:
y = x + 2 this polynomial has one zero.

Question 4.
How will you verify if a quadratic polynomial it has only one zero ?(Page No. 55)
Answer:
The linear polynomial ax + b, a ≠ 0 has exactly one zero namely the x co-ordinate of the point where the graph of y = ax + b intersects X-axis. Every linear polynomial in ax + b, a ≠ 0 has exactly one zero.

Question 5.
Write three quadratic polynomials that have no zeroes for x that are real numbers. (Page No. 55)
Answer:

1. y = 8x² – 6x + 1
2. y = x² – 8x + 17
3. y = x² – 4x + 5

These polynomials have no zeroes for x.

Question 6.
Find the zeroes of cubic polynomials
(i) -x³
(ii) x² – x³
(iii) x³ – 5x² + 6x without drawing the graph of the polynomial. (Page No. 57)
Solution:
i) p(x) = -x³.
p(x) = 0
-x³ = 0
x³ = 0
x = 0. So ‘0’ is the zero of the polynomial -x³.

ii) p(x) = x² – x³
= x²(1 – x)
p(x) = 0
x²(1 – x) = 0 ⇒ x² = 0 or 1 – x = 0
x = 0 (or) x = 1
0 & 1 are the zeroes of the polynomial x² – x³

iii) p(x) = x³ – 5x² + 6x
= x(x² – 5x + 6)
= x(x² – 3x – 2x + 6)
= x[x(x-3) – 2(x – 3)]
= x[(x – 3) (x – 2)]
p(x) =0
x[(x – 3) (x – 2)] = 0 ⇒ x = 0, x – 3 = 0
⇒ x = 3, x – 2 = 0 ⇒ x = 2.
∴ 0, 2 and 3 are the zeroes of the polynomial x³ – 5x² + 6x

Do This

Question 1.
Find the zeroes of the quadratic polynomials given below. Find the sum and product of the zeroes and verify relationship to the coefficients of terms in the polynomial .
(i) p(x) = x² – x – 6
(ii) p(x) = x² – 4x + 3
(iii) p(x) = x² – 4
(iv) p(x) = x² + 2x + 1. (Page No. 62)
Solution:
i) The given polynomial is x² – x – 6.
x² – x – 6 = x² – 3x + 2x – 6
= x(x – 3) +2(x – 3)
= (x + 2) (x – 3)
So, the value of x² – x – 6 is zero when x + 2 = 0 or x – 3 = 0 (i.e.,) when x = -2 or x = 3
Therefore, the zeroes of x² – x – 6 are -2 & 3
Now, sum of the zeroes = -2 + 3 = 1

Product of the zeroes = -2 × (3) = -6

ii) The given polynomial is x² – 4x + 3. (A.P. Mar. June 15)
x² – 4x + 3 = x² – 3x – x + 3
= x(x – 3) -1(x – 3)
= (x – 1) (x – 3)
So, the value of x² – 4x + 3 is zero when (x – 1) = 0 or (x – 3) = 0
(i.e.,) when x = 1 or x = 3
∴ The zeroes of x – 4x + 3 are 1 & 3.
Now sum of zeroes = 1 + 3 = 4

Product of the zeroes = 1 × 3 = 3

iii) The given polynomial is x² – 4
x² – 4 = (x)² – (2)² = (x + 2) (x – 2)
So, the value of x² – 4 is zero when (x + 2) = 0 or (x – 2) = 0
(i.e.,) when x = -2 or x = 2
∴ The zeroes of x² – 4 are -2 & 2.
Now, sum of the zeroes = -2 + 2 = 0

Product of the zeroes = (-2) × (2) = – 4

iv) The given polynomial is x² + 2x + 1.
x² + 2x + 1 = x² + x + x + 1
= x(x +1) +1 (x + 1)
= (x + 1) (x + 1)
= (x + 1)²
So, the value of x² + 2x + 1 is zero when (x + 1) = 0 = x + 1 = 0 (i.e.,) when x = -1 & x = -1.
∴ The zeroes of the polynomial are -1 & -1. Now, sum of the zeroes = -1 + (-1)
= -1 -1 = – 2

Question 2.
If α, β, γ are the zeroes of the given cubic polynomials, find the values as given in the table. (Page No. 66)

Solution:

Try This

Question 1.
i) Find a quadratic polynomial with zeroes -2 and 1/3. (Page No. 64)
Solution:
Let the quadratic polynomial be ax² + bx + c, a ≠ 0 & its zeroes be α & β.
Here α = -2; β = 1/3
sum of the zeroes = (α + β)
sum of the zeros = (α + β)
= (-2) + \(\frac{1}{3}\) = \(\frac{-6+1}{3}\) = –\(\frac{5}{3}\)
Product of the zeroes = αβ = (-2)(\(\frac{1}{3}\))
= –\(\frac{2}{3}\)
∴ The quadratic polynomial
ax² + bx + c is
k[x² – (α + β)x + αβ] where k is a constant.
= k[x² – \(\left(\frac{-5}{3}\right)\)x + \(\left(\frac{-2}{3}\right)\)]
We can put different values of k.
When k = 3 the quadratic polynomial will be 3x² + 5x – 2.

ii) What is the quadratic polynomial whose sum of zeroes is \(\frac{-3}{2}\) and the product of zeroes is -1 ? (Page No. 64) (A.P. June’15)
Solution:
Sum of zeroes = (α + β) = \(\frac{-3}{2}\)
Product of zeroes = αβ = -1
∴ The quadratic polynomial
ax² + bx + c is k[x² – (α + β)x + αβ]
We can put different values of k.
When k = 2 the quadratic polynomial will be 2x² + 3x – 2.