Question

5) A boy of height 1.5m is standing in front of a building of height 30m. As he
moves towards the building the angle of elevation to the top of building
increases from 30° to 60°. Find the distance moved by the boy.
​

Answer 1

Plz Mark me BRAINLIEST....!!! !!

Answer 2

Answer:

let AB=CD= EQ = 1.5 m

Let dist. covered by boy is BD=x

Let BD=AC=x

Let DQ=CE=y

In fig. PE=30-1.5=28.5m

Angles are 30° and 60°

In A PAE

tan 30 = PE/AE

1/√3=28.5/x+y

28.5√3=x+y -(1)

In ∆PCE

tan60=PE/CE

√3=28.5/y

y=28.5/√3

Put in (1) x+28.5/√3=28.5√4 x=(28.5×3-28.5)/√3

= (85.5-28.5)/V3 =57/√3

Rationalize =57/√3*√3/√3

=57√3/3

=19√3m