Srinivas

Students can practice 10th Class Maths Study Material Telangana Chapter 10 Mensuration Ex 10.3 to get the best methods of solving problems.

TS 10th Class Maths Solutions Chapter 10 Mensuration Exercise 10.3

Question 1.
An iron pillar consists of a cylindrical portion of 2.8 m height and 20 cm in di-ameter and a cone of 42 cm height sur-mounting it. Find the weight of the pillar if 1cm3 of iron weighs 7.5g. (AS4)
Solution:
Volume of the iron pillar = Volume of the cylinder + volume of the cone
Cylinder : Radius = \(\frac{\text { diameter }}{2}\)
= \(\frac{20}{2}\) = 10 cm
Height = 2.8 m
= 280 cm
Volume = πr²h
= \(\frac{22}{7}\) × 10 × 10 × 280
= 88,000 cm³

Cone : Radius ‘r’ = \(\frac{\text { diameter }}{2}\)
= \(\frac{20}{2}\) = 10 cm
Height ‘h’ = 42 cm
Volume = \(\frac{1}{3}\)πr²h
= \(\frac{1}{3}\) × \(\frac{22}{7}\) × 10 × 10 × 42
= 4,400 cm³.
Total volume = 88,000 + 4,400
= 92,400 cm³
∴ Total weight of the pillar at a weight of 7.5g per 1 cm³ = 92,400 × 7.5
= 6,93,000 gms
= \(\frac{6,93,000}{1,000}\) kg = 693 kg

Question 2.
A toy is made in the form of hemisphere surmounted by a right cone whose cicular, base is joined with the plane surface of hemisphere. The radius of the base of the cone is 7cm and its volume is \(\frac{3}{2}\) of hemisphere. Calculate the height of the cone and surface area of the toy correct to 2 places of decimal (take π = 3\(\frac{1}{7}\)) (AS4)
Solution:
Radius of the base of the cone(r) = 7 cm
Radius of hemisphere = 7cm
Volume of hemisphere = \(\frac{2}{3}\) πr³
= \(\frac{2}{3}\) × \(\frac{22}{7}\) × 7 × 7 × 7cm³
Volume of the cone
= \(\frac{3}{2}\) × volume of the hemisphere (given)
= \(\frac{3}{2}\) × \(\frac{2}{3}\) × \(\frac{22}{7}\) × 7 × 7 × 7
= 1078

Volume of cone = \(\frac{1}{3}\) πr²h = 1,078
= \(\frac{1}{3}\) × \(\frac{22}{7}\) × 7 × 7 × h = 1078
∴ h = 1,078 × \(\frac{3}{1}\) × \(\frac{7}{22}\) × \(\frac{1}{7}\) × \(\frac{1}{7}\) = 21cm
l² = h² + r² = 21² + 7²
= 441 + 49 = 490
l = \(\sqrt{490}\) = \(\sqrt{49 \times 10}\)
= 7\(\sqrt{10}\) = 7 × 3.16 = 22.12 cm
Surface area of the toy = curved surface area of the cone + curved surface area of the hemisphere = πrl + 2 πr²
= \(\frac{22}{7}\) × 7 × 7 × \(\sqrt{10}\) + 2 × \(\frac{22}{7}\) × 7 × 7
= \(\frac{22}{7}\) × 7 [7\(\sqrt{10}\) + 14]
= 22 × (22.12 + 14)
= 22 × 36.12 = 794.64 cm²

Question 3.
Find the volume of the largest right circular cone that can be cut out of a cube whose edge is 7cm. (AS4)
Solution:
The base of the largest right circular cone will be the circle inscribed in a face of the cube and its height will be equal to an edge of the cube.
∴ r = Radius of the base of the cone
= \(\frac{7}{2}\) cm (∴ edge = 7cm)
h = height of the cone = 7cm
Hence, volume of the cone
= \(\frac{1}{3}\) πr²h
= \(\frac{1}{3}\) × \(\frac{22}{7}\) × \(\frac{7}{2}\) × \(\frac{7}{2}\) × 7
= \(\frac{539}{6}\) cm³ = 89.83 cm³

Question 4.
A cylindrical tub of radius 5 cm and length 9.8 cm is full of water. A solid in the form of right circular cone mounted on a hemisphere is immersed into the tub. The radius of the hemisphere is 3.5 cm and height of cone outside the hemisphere is 5 cm. Find the volume of water left in tub (Take π = \(\frac{22}{7}\)) (AS4)
Solution:
Radius of the cylinder(r) = 5 cm
Height of the cylinder (h) = 9.8 cm
Volume of the cylinder = πr²h

= \(\frac{22}{7}\) × 5 × 5 × 9.8
= 770 cm³.
Radius of hemi-sphere(r) = 3.5 cm
Volume of the hemi-sphere = \(\frac{2}{3}\) πr³
= \(\frac{2}{3}\) × \(\frac{22}{7}\) × 35 × 35 × 35
= \(\frac{539}{6}\) cm³

Radius of the base of the cone(r) = 3.5 cm
Height of the cone = 5 cm
∴ Volume of the cone = \(\frac{1}{3}\)πr²h
= \(\frac{1}{3}\) × \(\frac{22}{7}\) × 3.5 × 3.5 × 5
= \(\frac{385}{6}\) cm³
Total volume of the solid = volume of the hemisphere + volume of the cone
= \(\frac{539}{6}\) + \(\frac{385}{6}\) = \(\frac{924}{6}\) = 154 cm²
The volume of water left in the tub = volume of the cylindrical tube – volume of the solid immersed
= 770 – 154 = 616 cm³

Question 5.
In the adjacent figure, the height of a solid cylinder is 10 cm and diameter is 7 cm. Two equal conical holes of radius 3 cm and height 4 cm are cut off as shown in figure. Find the volume of the remaining solid. (AS4)

Solution:
Radius of the solid cylinder (r) = \(\frac{7}{2}\) cm
(∴ diameter = 7cm)
Height of the solid cylinder(h) = 10 cm
Volume of the solid cylinder = πr²h
= \(\frac{22}{7}\) × \(\frac{7}{2}\) × \(\frac{7}{2}\) × 10 = 385 cm³
Radius of concial hole(r) = 3 cm
Height(depth) of the conical hole(h) = 4 cm
Volume of the conical part = \(\frac{1}{3}\) pr²h
= \(\frac{1}{3}\) × \(\frac{22}{7}\) × 3 × 3 × 4
Total volume of two conical holes
= 2 × \(\frac{1}{3}\) × \(\frac{22}{7}\) × 3 × 3 × 4
= \(\frac{528}{7}\) = 75.43 cm³
Hence, volume of the remaining solid
= [Volume of the Cylindrical part] – [Total volume of two conical holes]
= 385 – 75.43
= 309.57cm³

Question 6.
Spherical Marbles of diameter 1.4 cm are dropped into a cylindrical beaker of diameter 7cm, which contains some water. Find the number of marbles that should be dropped into the beaker, so that water level rises by 5.6 cm. (AS4) Solution:
Rise in the water level is seem in cylinderial shape of Radius = Beaker radius

= \(\frac{\text { diameter }}{2}\) = \(\frac{7}{2}\) = 3.5 cm
Height ‘h’ of the rise = 5.6 cm
∴ Volume of the ‘water rise’ = πr²h
= \(\frac{22}{7}\) × 3.5 × 3.5 × 5.6
= \(\frac{22 \times 12.25 \times 5.6}{7}\) = 215.6
Volume of each marble dropped
= \(\frac{4 \pi r^3}{3}\)
Where radius r = \(\frac{\mathrm{d}}{2}\) = \(\frac{1.4}{2}\) = 0.7 cm
∴ V = \(\frac{4}{3}\) × \(\frac{22}{7}\) × 0.7 × 0.7 × 0.7
≈ 1.4373 cm³
∴ Volume of the ‘rise’ = Total volume of the marbles.
Let the number of marbles be ‘n’ then
n × volume of each marble = Volume of the rise
n × 1.4373 = 215.6
n = \(\frac{215.6}{1.4373}\) = 150
∴ Number of marbles = 150

Question 7.
A pen stand is made of wood in the shape of cuboid with three conical depression to hold the pens. The dimensions of the cuboid are 15 cm by 10cm by 3.5cm. The radius of each of the depression is 0.5 cm and the depth is 1.4 cm. Find the volume of wood in the entire stand. (AS4)

Solution:
The dimensions of the cuboid are
15 cm × 10 cm × 3.5 cm
∴ Volume of the ‘cuboid’ = \(\frac{15 \times 10 \times 7}{2}\)
= 525 cm³
Radius of the conical depression (r) = 0.5 cm
Depth of conical hole (h) = 1.4 cm
= \(\frac{14}{10}\) cm
Volume of one conical hole
= \(\frac{1}{3}\) × \(\frac{22}{7}\) × \(\frac{1}{2}\) × \(\frac{1}{2}\) ×\(\frac{14}{10}\)
Volume of three conical holes
= 3 × \(\frac{1}{3}\) × \(\frac{22}{7}\) × \(\frac{1}{2}\) × \(\frac{1}{2}\) × \(\frac{14}{10}\)
= \(\frac{11}{10}\) = 1.1 m³
∴ The volume of wood in the entire stand
= (Volume of the cuboid) – (volume of 3 conical holes)
= 525 – 1.1 = 523.9cm³.

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 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 10th Class Maths Textbook SSC Solutions Telangana Chapter 13 Probability Ex 13.1 to get the best methods of solving problems.

TS 10th Class Maths Solutions Chapter 13 Probability Exercise 13.1

Question 1.
Complete the following statements : (AS₃)
i) Probability of an event E + Probability of the event ‘not E’ = ____________
ii) The probability of an event that cannot happen is _________ Such an event is called __________
iii) The probability of an event that is certain to happen is _________ Such an event is called _________
iv) The sum of the probabilities of all the elementary events of an experiment is _________
v) The probability of an event is greater than or equal to _________ and less than or equal to.
Solution:
i) 1
ii) 0, impossible event
iii) 1, sure (or) certain event
iv) 1
v) 0, 1

Question 2.
Which of the following experiments have equally likely outcomes ? Explain. (AS₃)
i) A driver attempts to start a car. The car starts or does not start.
Solution:
Equally likely. Since both have the some probability = \(\frac{1}{2}\)

ii) A player attempts to shoot a basket ball. She! he shoots or misses the shot.
Solution:
Equally likely. Since both have the same probability = \(\frac{1}{2}\)

iii) A trial ¡s made to answer a true-false question. The answer is right or wrong.
Solution:
Equally likely. Since both have the same probability = \(\frac{1}{2}\)

iv) A baby is born. It is a boy or a girl.
Solution:
Equally likely. Since both the events have the same probability = \(\frac{1}{2}\)

Question 3.
If P(E) = 0.05, what is the probability of “not E” ? (AS₁)
Solution:
Given that P(E) = 0.05
Probability of “not E” is denoted by \(\overline{\mathrm{E}}\)
P(\(\overline{\mathrm{E}}\)) = 1 – P(E)
= 1 – 0.05
= 0.95

Question 4.
A bag contains lemon flavoured candies only. Malini takes out one candy without looking into the bag. What is the probability that she takes out.
i) an orange flavoured candy ? (AS₄)
ii) a lemon flavoured candy ?
Solution:
i) The bag contains lemon flavoured candies only.
(i.e.,) It does not contains orange flavoured candies.
So, the probability that she takes out an orange flavoured candy is ‘O’.

ii) Probability that she takes out lemon flavoured candy is 1 because the bag contains lemon flavoured candies only.

Question 5.
Rahim removes all the hearts from the cards. What is the probability of she takes out
i) Getting an ace from the remaining pack. (AS₄)
ii) Getting a diamond.
iii) Getting a card that is not a heart.
iv) Getting the Ace of hearts.
Solution:
All the hearts are taken out of 52 cards.
Therefore, the remaining cards will be 52 – 13 = 39
i) The probability of getting an ace from the remaining pack.
= \(\frac{\text { Number of outcomes favourable to the event }}{\text { Total number of all possible outcomes }}\)
= \(\frac{4}{52}\) = \(\frac{1}{13}\)

ii) The probability of picking out a diamond
= \(\frac{\text { Number of outcomes favourable to the event }}{\text { Total number of all possible outcomes }}\)
= \(\frac{13}{39}\) = \(\frac{1}{3}\)

iii) The probability of getting a card that is not a heart.
= \(\frac{\text { Number of outcomes favourable to the event }}{\text { Total number of all possible outcomes }}\)
= \(\frac{39}{39}\) = 1

iv) The probability of getting the Ace of hearts.
= \(\frac{\text { Number of outcomes favourable to the event }}{\text { Total number of all possible outcomes }}\)
= \(\frac{0}{39}\) = 0

Question 6.
It is given that in a group of 3 students, the probability of 2 students not having the same birthday is 0.992. What is the probability that the 2 students have the same birthday ? (AS₄)
Solution:
Probability that the 2 students have the same birthday.
= 1 – Probability that the 2 students do not have the same birthday.
= 1 – 0.992 = 0.008

Question 7.
A die is rolled once. Find the probability of getting
(i) a prime number ;
(ii) a number lying between 2 and 6 ;
(iii) an odd number. (AS₁, AS₄)
Solution:
Suppose we roll a die once. The possible out-comes are 1, 2, 3, 4, 5 and 6. Each number has the same possibility of showing up. So the equally likely outcomes of rolling a dice are 1, 2, 3, 4, 5 and 6.

i) Let E be the event of getting a prime number. Then the outcomes favourable to E are 2, 3 and 5 (prime numbers).
Therefore, the number of outcomes favourable to E is 3.
So, P(E)
= \(\frac{\text { No. of outcomes favourable to E}}{\text { No. of all possible outcomes }}\)
= \(\frac{3}{6}\) = \(\frac{1}{2}\)

ii) The numbers lying between 2 and 6 are 3, 4 and 5.
Let E be the event of getting a number lying between 2 and 6.
Then, the outcomes favourable to E are 3, 4 and 5.
Therefore the number of outcomes favourable to E is 3.
So, P (E)
= \(\frac{\text { No. of outcomes favourable to E}}{\text { No. of all possible outcomes }}\)
= \(\frac{3}{6}\) = \(\frac{1}{2}\)

iii) The odd numbers in the first 6 natural numbers is 1, 3 and 5.
Let E be the event of getting an odd number.
Then, the outcomes favourable to E are 1, 3 and 5.
Therefore, the number of outcomes favourable to E is 3.
So, P(E)
= \(\frac{\text { No. of outcomes favourable to E }}{\text { No. of all possible outcomes }}\)
= \(\frac{3}{6}\) = \(\frac{1}{2}\)

Question 8.
What is the probability of selecting out a red king from a deck of cards ?
Solution:
A deck has 52 cards.
Since the total number of cards is 52.
Number of all possible outcomes = 52
Let E be the event of getting a red king.
Then, the outcomes favourable to E are king of diamond and heart. Therefore, the number of outcomes favourable to E is 2.
So, P(E)
= \(\frac{\text { No. of outcomes favourable to E}}{\text { No. of all possible outcomes }}\)
= \(\frac{2}{52}\) = \(\frac{1}{26}\)

Question 9.
Make 5 more problems of this kind using die, cards or birthdays and discuss with friends and teacher about their solutions. (AS₃)
Solution:

1. A die is thrown once. Find the probability of getting a number less than 3.
2. From a well shuffled pack of cards is drawn at random. Find the probability of getting a black queen.
3. A die is thrown once. What is the probability of
   a) Getting a number greater than 2 ?
   b) Getting an even number ?
   c) Getting a number between 3 and 6 ?
4. Two coins are tossed simultaneously. Find the probability of getting exactly one head.
5. Raghu and Siva are friends. What is the probability that both will have
   1. Different birthdays ?
   2. The same birthday ? (ignoring a leap year)
6. The king, queen and jack of clubs are removed from a deck of 52 playing cards and the remaining cards are shuffled. A card is drawn from the remaining cards. Find the probability of getting a card of
   1. Heart
   2. Queen

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

TS 10th Class Maths Solutions Chapter 14 Statistics InText Questions

Think – Discuss

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

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

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

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

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

Do This

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

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

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

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

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

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

Think – Discuss

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

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

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

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

TS 10th Class Maths Solutions Chapter 14 Statistics Exercise 14.2

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

Find the mode and the mean of the data given above. Compare and interpret the two measures of central tendency. (AS₃, AS₄)
Solution:
Maximum number of patients joined in the age group 35 – 45
∴ Modal class is 35 – 45.
Lower limit of the modal class ‘l’ = 35
class size, h = 10
Frequency of modal class. f₁ = 23
Frequency of the class preceding the modal class f₀ = 21
Frequency of the class succeding the modal class f₂ = 14

∴ Mode = l + \(\frac{\left(f_1-f_0\right)}{2 f_1-f_0-f_2}\) × h
= 35 + \(\left[\frac{23-21}{2 \times 23-21-14}\right]\) × 10
= 35 + \(\left[\frac{2}{46-35}\right]\) × 10
= 35 + \(\frac{2}{11}\) × 10 = 35 + 1.81818 ….
= 36.8 years

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

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

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

Determine the modal life times of the components.
Solution:

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

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

Solution:

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

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

Solution:

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

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

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

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

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

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

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