Question

The angle of depression of a car standing on the ground from the top of a 66 m tower is 30 degree find the distance of the car from the base of the tower

Answer 1

Given :-

• Angle of depression of car is 30°

• Height of tower is 66 m

To Find :-

• Distance between car and base of tower.

Solution :-

Let's assume that AB is tower having 66 m height . Here angle B is 90° and C is the point on the ground where car is standing .

Angle ACB is 30° here .

$\underline{\boxed{\sf{\tan \theta = \frac{Perpendicular}{base} }}}$

• $\sf{Perpendicular= 66m }$

• $\sf{Base= xm }$

• $\sf{\theta = 30}$

Putting these values in the formula mentioned .

$\sf{\implies \tan 30 = \frac{66}{x}}$

$\sf{\implies \frac{1}{\sqrt{3}} = \frac{66}{x} }$

$\sf{\implies x = 66 \sqrt{3} }$

So the distance between base of tower and car is 66√3 m .

Answer 2

${\underline{\sf{Question}}}$

The angle of depression of a car standing on the ground from the top of a 66 m tower is 30 degree find the distance of the car from the base of the tower .

${\underline{\sf{Solution}}}$

Let PQ denotes the tower of length 66 m and R denotes the position of car.

and The angle of depression of a car standing on the ground from the top of tower is 30°

We have to find the distance of the car from the base of the tower .i.e length of RQ .

From the figure :

In ∆ PQR

$\sf\:\tan\theta=\frac{PQ}{RQ}$

$\sf\:\implies\:\tan\theta=\frac{66}{RQ}$

$\sf\:\implies\:tan30\degree=\frac{66}{RQ}$

$\sf\:\implies\dfrac{1}{\sqrt{3}}=\frac{66}{RQ}$

⇒RQ = 66√ 3

⇒RQ = 66× 1.732

⇒RQ = 114.312 m

Thus, the distance of a car from the base of the tower is 114.312 m.