Question

the angle of elevation of a tower from a point on ground,which is 60m away from foot of tower is 60° Find the height of tower
​

Answer 1

In ΔADC

tan30

o

=

AC

DC

3

1

=

d

h

........(1)

d=

3

h

→d=

3

×10

3

=30m

In ΔBCD

tan60

o

=

BC

DC

3

=

d−20

h

.......(2)

on solving equation (1) & (2)

3

=

3

h−20

h

⇒3h−20

3

=h

2h=20

3

h=10

3

m

height of tower.