Question

1.5 m tall boy is standing at some distance from a 30 m tall building The ang
elevation from his eyes to the top of the building increases from 30 to 60°
towards the building. Find the distance he walked towards the building​

Answer 1

Answer:

REF.Image

AB= 30-1.5=28.5 cm

In ΔABD:−tan30∘=BDAB

⇒BD=(28.5)3

In ΔABC:−tan60∘=BCAB

⇒BC=3(28.5)

CD= BD-BC

=28.5(3−31)

=28.5(32)=32.91m