Question

3) The angle of elevation of a top of the tower from two points at a distance of 3m and 12m
from the base of the tower and in the same straight line with it are complementary. Find the
height of the tower.​

Answer 1

Answer:

Given,

the angle of elevation of the top of the tower from two points P & Q is at a distance of a & b.

Also given, to prove that the tower

height=

ab

(∵ complementary angle =(90

o

−θ))

From ΔABP

tanθ=

BP

AB

=

a

AB

……..(1)

From ΔABQ

tan(90−θ)=

BQ

AB

(∵tan(90−θ)=cotθ)

(cotθ=

tanθ

1

)

We get,

cotθ=

AB

BQ

=

AB

b

……..(2)

by equation (1) & (2) we get,

a

AB

=

AB

b

⇒AB

2

=ab⇔AB=

ab

∴AB=height=

ab

Hence proved.

solution