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 meters 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 ).

Answer 1

Answer:

apply the formula tan60=

$\sqrt{3}$

apply formula tan 30 =

$1 \div \sqrt{3}$

Answer 2

The height of tower is 129.9 meters , proved .

Step-by-step explanation:

Given as :

The angle of elevation of a tower from a point on the same level as the foot of the tower is 30°.

Further 150 meters , towards tower the elevation changes to 60°

Let The height of tower = H meters

According to question

From figure

In ΔOBA

Tan angle = $\dfrac{perpendicular}{base}$

Tan 60°  =  $\dfrac{OA}{OB}$

Or, √3 = $\dfrac{H}{x}$

∴   H = √3 x                               ..........1

Again

In ΔOCA

Tan angle = $\dfrac{perpendicular}{base}$

Tan 30°  =  $\dfrac{OA}{OC}$

Or, $\dfrac{1}{\sqrt{3} }$ = $\dfrac{H}{150+x}$

∴  150 + x = √3 H                .......2

From eq 1 and eq 2

150 + x = √3 × √3 x

Or, 150 = 3 x - x

Or, 2 x = 150

∴       x = $\dfrac{150}{2}$

i.e      x = 75 meters

Now, Put the value of x in eq 1

H = √3 x  

Or,H = √3 × 75

So, Height of tower = H = 75√3  = 129.9 meters

Hence , The height of tower is 129.9 meters , proved . Answer