P and Q are two points observed from the top of a building 10√3 m high. If the angles of depression of the points are complementary and PQ = 20m, then the distance of P from the building is

Let the unknown angle of depression be x.Since the angles of depression are complementary, the other angle is (90-x)
$$\angle ROQ = x$$ since $$\triangle$$ ROQ is a right angled triangle.
$$h = 10 \sqrt{3}$$
Let RQ = y metres
tan x = $$\frac{OR}{RP} = \frac{RQ}{OR}$$
$$\frac{OR}{RP}=\frac{h}{y+20}$$
$$\frac{RQ}{OR} = \frac{y}{h}$$
$$\frac{h}{y+20} = \frac{y}{h}$$
$$\frac{10\sqrt{3}}{y+20} = \frac{y}{10\sqrt{3}}$$
$$y(y+20)= 300$$
Solving for y we get y = 10 m.
RP = 20 +10 m = 30m is the distance of P from the building.
Option C is the correct answer.