sir this is the important question from the book -ML aggarwal( avichal publication) class 10th , chapter20 , heights and distances

it is given that From the top of a church spire 96 m high,

the angles of depression of two vehicles on a road,

at the same level as the base of the spire and on the same side of it are x0 and y0,

where tan x0 = ¼ and tan y0 = 1/7.

Calculate the distance between the vehicles

question no 20 , heights and distances , ICSE board

Consider CH as the height of the church

A and B are two vehicles which make an angle of depression x

^{0}and y^{0}from CTake AH = x and BH = y

In a right triangle CBH

tan x

^{0}= CH/AH = 96/ySubstituting the values

1/4 = 96/y

So we get

y = 96 × 4 = 384 m

In right triangle CAH

tan y

^{0}= CH/AH = 96/xSubstituting the values

1/7 = 96/x

So we get

x = 96 × 7 = 672 m

Here

AB = x – y

AB = 672 – 384

AB = 288 m