Question

The angle of elevation of the top of a 78m high tower from a point A on the ground is 30°. Find

the distance of the point A from the foot of the tower.​

Answer 1

In ΔADC

tan30° = AC DC

31 = dh........(1)

d= 3 h

→d= 3 ×10 3

=30m

In ΔBCD

tan60° = BC DC

3 = d−20h .......(2)

on solving equation (1) & (2)

3 = 3 h−20h

⇒3h−20 =h

2h=20 3

h=10

3 mheight of tower.

Answer 2

Answer:

The distance of point A from the foot of the tower = 78$\sqrt{3\\}$ m .

Step-by-step explanation:

Tan A =  $\frac{height of the tower}{distance from point A }$

Tan 30 = $\frac{78}{x}$

$\frac{1}{\sqrt{3} }$ = $\frac{78}{x}$

By cross multiplying,

x = 78$\sqrt{3}$