Question

from a point 20m away from the foot of a tower , the angle of elevation of the top of the tower is 60degree .find the height of the tower . at what distance from the foot of the tower the angle of elevation will be 30degree?

Answer 1

Let AB be the tower,and from D,the angle of elevation of the tower is 30°.
In right ΔABC,
AB/BC=tan 60°⇒AB/20=√3⇒AB=20√3 m
∴Height of the tower=20√3 m.

In right ΔABD,
AB/BD=tan 30°⇒20√3/BD=1/√3⇒BD=60 m
∴At a distance of 60 m from the foot of the tower,the angle of elevation of the tower is 30°.

Answer 2

Answer:

Height of the tower is 20 root3

and,BD =60m

Step-by-step explanation:

Let AB be the tower

<ADC=30°

in ABC

AB/BC=tan60

AB/20=root3

AB= 20 root3(height of the tower)

Now in ABD

AB/BD=tan30

20root3/BD=1/root3

BD=60m