Question

On one side of the road there is a building and on
other side of the road is a tower. From the top of the tower
the angle of depression of the roof and base of the building is
45° and 60° respectively. If the height of the building is 10 m,
then find the height of the tower.
4​

Answer 1

Answer:

Let AB is a building of height 12m. CE is a tower. Let ED=h m.

The angles of depression from point E of top of tower at the roof and base on building are 45

0

and 60

0

respectively. Now

∠XEA=∠EAD=45

0

(Alternate angle)

∠XEB=∠EBC=60

0

(Alternate angle)

Let BC=x and ED=h m

AB=CD=12m

From right angled ΔEAD,

tan45°=

AD

ED

l=

x

h

h=x …(i)

From right angled ΔEBC,

tan60°=

BC

h+12

3

=

x

h+12

=

h

h+12

(Put the value of x from equation)

3

h=h+12

3

h–h=12

h[1.732–1]=12

h=

0.732

12

=16.393m

Hence, height of tower =EC=CD+ED

=12+16.393

=28.393m

solution