Question

The angle of elevation of a tower from a point on the same level as the foot of the tower is 30°. On advancing 150 metres towards the foot of the tower, the angle of elevation of the tower becomes 60°. Show that the height of the tower is 129.9 metres (Use √3=1.732)

Answer 1

Answer:

The height of the tower is 129.9 m

Step-by-step explanation:

Let AB be the height of the tower (h).

Given :  

Angle of elevation of the top of the tower (θ), ∠ADB = 30°

He walks CD = 150 m towards the foot of the tower,then the Angle of elevation of  the top of the tower ,(θ), ∠ACB = 60°

Let CB = x m

In right angle triangle, ∆ABC,

tan θ  = P/ B

tan 60° = AB/BC

√3 = h/x

x = h/√3 ……...…..(1)

In right angle triangle, ∆ABD

tan θ  = P/ B

tan 30° = AB/BD

1/√3 = AB/(BC + CD)

1/√3 = h/(x + 150)

√3h = x + 150

√3h = h/√3 + 150

[From eq 1]

√3h - h/√3 = 150

(√3 ×√3 h - h)/√3 = 150

(3h - h)/√3 = 150

2h = 150√3

h = (150√3)/2

h = 75√3

h = 75 × 1.732

[√3 = 1.732]

h = 129.9  m

Hence, the height of the tower is 129.9 m

HOPE THIS ANSWER WILL HELP YOU…

Answer 2

SOLUTION

Use trigonometric ratios in the ∆ABD,

$tan60 \degree = \frac{h}{a} \\ \\ = > \sqrt{3} = \frac{h}{a} \\ \\ = > h = \sqrt{3a} .............(1)$

Similarly using trigonometric ratios in the triangle ABC, we get

$tan30 \degree = \frac{h}{a + 150} \\ \\ = > \sqrt{3h} = 150 + a..............(2)$

Use equation (1) in equation (2), we get

a= 75..........(3)

Putting the value of a in equation (1) to get the value of h

h= 75√3

=) h= 75(1.732)

=) h= 75× 1.732

=) h= 129.9 m

Therefore, the height of the tower is 129.9m.

hope it helps ☺️