Question

The angle of depression of a car parked on the road from the top of a 150 m high tower is 30º. The distance of the car from the tower (in metres) is
(a)50√3
(b)150√3
(c)150√2
(d)75

Answer 1

Answer:

The distance of the car from the tower is 150√3 m  .

Among the given options option (b) 150√3 m  is correct.

Step-by-step explanation:

Given : The height of the tower , AB = 150 m

∠ACB ,(θ) = ∠CAD = 30° (alternate angles)

Let the distance of the car from the tower is BC .

In right angle ∆ ABC ,  

tan θ = P/B  

tan 30°  = AB/BC

1/√3 = 150/BC

BC = 150√3 m  

Hence, the distance of the car from the tower is 150√3 m  .

HOPE THIS ANSWER WILL HELP YOU…

Answer 2

$\huge\bigstar\mathfrak\red{\underline{\underline{SOLUTION:}}}$

We have a high tower whose height (AC)= 150cm

Angle of depression= 30°

Here given ∠ of depression ∠DCB=30°

We know,∠DCB= ∠ABC (Alternate angle)

Let, distance of the car from tower AB= x m.

In ∆ABC,

$tan 30 \degree = \frac{AC}{AB} \\ \\ = > \frac{1}{ \sqrt{3} } = \frac{150}{x} \\ \\ = > x = 150 \sqrt{3} m$

Hope it helps ☺️