1. 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
Write Comment
Comments
By: anil on 05 May 2019 01.58 am
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 $$ riangle$$ ROQ is a right angled triangle.
$$h = 10 sqrt{3}$$
Let RQ = y metres
tan x = $$frac{OR}{RP} = frac{RQ}{OR}$$
$$frac{OR}{RP}=frac{h}{y+20}$$
$$frac{RQ}{OR} = frac{y}{h}$$
$$frac{h}{y+20} = frac{y}{h}$$
$$frac{10sqrt{3}}{y+20} = frac{y}{10sqrt{3}}$$
$$y(y+20)= 300$$
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.
Terms And Service:We do not guarantee the accuracy of available data ..We Provide Information On Public Data.. Please consult an expert before using this data for commercial or personal use
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 $$ riangle$$ ROQ is a right angled triangle.
$$h = 10 sqrt{3}$$
Let RQ = y metres
tan x = $$frac{OR}{RP} = frac{RQ}{OR}$$
$$frac{OR}{RP}=frac{h}{y+20}$$
$$frac{RQ}{OR} = frac{y}{h}$$
$$frac{h}{y+20} = frac{y}{h}$$
$$frac{10sqrt{3}}{y+20} = frac{y}{10sqrt{3}}$$
$$y(y+20)= 300$$
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.