10th mathematics exercise questions with answers

# Trigonometry Exercise 13.5 – Class 10

### Trigonometry Exercise 13.5 – Questions:

1. Find the value of ‘x’:

(i) (ii) (iii) (iv) (v) II.

1. A tall building casts a shadow of 300 m long when the sun’s altitude (elevation) is 30˚. Find the height of the tower.
2. From the top o a building 50√3 m high, the angle of depression of an object on the ground is observed to be 45˚. Find the distance of the object from the building.
3. A tree is broken over by wind forms a right angled triangle with the ground. If the broken part makes an angle of 60˚ with the ground and the top of the tree is now 20m from its base, how tall was the tree?
4. The angle of elevation of the top of a flag-post from a point on a horizontal ground is found to be 30˚. On walking 6 m towards the post, the elevation increased by 15˚. Find the height of the flag-post.
5. The angles of elevation of the top of the cliff as seen from the top and bottom of a building are 45˚ and 60˚ respectively. If the height of the building is 24 m, find the height of the cliff.
6. From the top a building 16 m high, the angular elevation of the top of a hill is 60˚ and the angular depression of the foot of the hill is 30˚. Find the height of the hill.
7. Find the angle of depression if an observer 150 m tall looks at the tip of his shadow which is 150√3 cm from his foot.
8. From a point 50 m above the ground the angle of elevation of a cloud is 30˚ and the angle of depression of its reflection in water is 60˚. Find the height of the cloud above the ground.

## Trigonometry Exercise 13.5 – Solutions:

1. Find the value of ‘x’:

(i) Solution:

In ∆ABC , ∠ACB = 45˚

tanθ  =  AB/BC

tan 45˚ = x/60

1 = x/60

x = 60 m

(ii) Solution:

In ∆PRQ , ∠RQP = 60˚

tanθ  =  RP/PQ

tan 60˚ = 90/x

√3 = 90/x

√3 x = 90

x = 30√3 units

(iii) Solution:

In ∆LKM , ∠KLM = 30˚

tanθ  =  KM/LM

tan 30˚ = x/100

1/√3 = x/100

x√3 = 100

x = 100/√3 units

(iv) Solution:

In ∆XYZ , ∠XZY = 45˚

cosθ  =  YZ/XZ

cos 45˚ = x/100

1/√2 = x/100

x√2 = 100

x = 100/√2 units

(v) Solution:

In ∆DEF , ∠DEF = x

tanθ  =  DF/FE

tan x = 75/75

tan x = 1

We know, tan 45˚ = 1

tan x = tan 45˚

x = 45˚

II.

1. A tall building casts a shadow of 300 m long when the sun’s altitude (elevation) is 30˚. Find the height of the tower.

Solution: In ∆ABC , ∠ACB = 30˚

tanθ  =  AB/BC

tan 30˚ = x/300

1/√3 = x/300

x√3 = 300

x = 300/√3 = 100√3

1. From the top of a building 50√3 m high, the angle of depression of an object on the ground is observed to be 45˚. Find the distance of the object from the building.

Solution: Let PQ be the height f the building, PQ = 50√3, QR be the distance between the building and object.

Angle of depression = 45˚

Since, PM||QR, ∠MPR = ∠PRQ (alternate angles)

∠MPR = 45˚

Therefore, ∠PRQ = 45˚

In ∆PQR , ∠MPR = 45˚

tanθ  =  PQ/QR

tan 45˚ = 50√3/QR

1 = 50√3/QR

QR = 50√3 m

1. A tree is broken over by wind forms a right angled triangle with the ground. If the broken part makes an angle of 60˚ with the ground and the top of the tree is now 20m from its base, how tall was the tree?

Solution: Let AC is the full length of the tree. AB is broken and  bends as BD. Therefore AB = BD.

Given, CD = 20 m and ∠BDC = 60˚

In ∆BCD , ∠BDC = 60˚

tanθ  =  BC/CD

tan 60˚ = 20/CD

√3 = BC/20

20√3 = BC

cosθ = CD/BD

1/2 = 20/BD

40 = BD

To find the full length of the tree , AC = BC + BD(since  AB = BD)

= 20√3 + 40

= 20(√3 + 2) m

1. The angle of elevation of the top of a flag-post from a point on a horizontal ground is found to be 30˚. On walking 6 m towards the post, the elevation increased by 15˚. Find the height of the flag-post.

Solution: Given ∠BAC = 30˚, ∠BDC = 45˚

tan30˚ = BC /AD + DC

1/√3 = BC/6+x

√3 .BC = 6 + x

BC  = 6+x/√3 ………….(1)

tan45˚ = BC/DC

1 = BC/x

BC = x…………..(2)

Substitute (2) in (1),

x = 6 + x/√3

√3.x = 6 + x

6 = √3.x – x

6 = x(√3 – 1)

x = 6/(√3 – 1)

1. The angles of elevation of the top of the cliff as seen from the top and bottom of a building are 45˚ and 60˚ respectively. If the height of the building is 24 m, find the height of the cliff.

Solution: Height od the building i.e., AD = BC = 24 m. Similarly, AB = DC,

In ∆MCD , ∠MDC = 60˚

tan θ = MC/DC

tan60˚ = 24+x/DC

√3 = 24+x/DC

DC.√3 = 24 + x

DC = 24+x/√3

In ∆AMB , ∠MAB = 45˚, MB = x

tan θ = MB/AB

tan45˚ = x/(24+x/√3)

1 = x√3/24+x

√3x = 24 + x

√3x – x = 24

(√3 – 1)x = 24

x = 24/√3 – 1

Height of the cliff, MC = MB + BC = (24 + 24/√3 – 1) m

1. From the top a building 16 m high, the angular elevation of the top of a hill is 60˚ and the angular depression of the foot of the hill is 30˚. Find the height of the hill.

Solution: In ∆DCE , ∠CED = 30˚, DE = x

tan θ = CD/DE

tan30˚ = 16/x

1/√3 = 16/x

x = 16√3

In ∆ABC , ∠BCA = 60˚, BC = DE = x

tan θ = AB/BC

tan 60˚ = AB/x

√3 = AB/16√3

16×3 = AB

48 = AB

Therefore, the height of the hill = AB + BE = 48 + 16 = 64 m

1. Find the angle of depression if an observer 150 m tall looks at the tip of his shadow which is 150√3 cm from his foot.

Solution: In ∆ABC , AB = 150 cm and BC = 150√3

tan θ = AB/BC = 150/150√3

tan θ = 1/√3

We know, tan30˚ = 1/√3

Therefore, tan θ = tan30˚

Hence, θ = 30˚

1. From a point 50 m above the ground the angle of elevation of a cloud is 30˚ and the angle of depression of its reflection in water is 60˚. Find the height of the cloud above the ground.

Solution: Let AC = h

BE = CD = 50 m

A’D = AD = 50 + h

In ∆ABC , ∠ABC = 60˚,

tan θ = AC/BC = h/BC

tan 30˚ = h/BC

1/√3 = h/BC

√3.h = BC

Now, in ∆A’CB ,

tan60˚ = CA’/BC = CD+DA’/BC

√3 = 50+50+h/√3h = 100+h/√3h

3h = 100 + h

2h = 100

h = 50 m

Thus, height of the cloud from the surface of lake = AD+A’D = 50 + 50  = 100m