Question

a tower is 64 m high.a man standing erect at a distance of 36 m from the tower observe the angle elevation of the top of the tower is 60 degree.find the height of the man.

Answer 1

hope it will be helpful to you

Answer 2

Concept:

The term angle of elevation denotes the angle from the horizontal upward to an object. An observer’s line of sight would be above the horizontal.

$tan\theta =\frac{opposite}{adjacent},\ tan60=\sqrt{3}$

Given:

It is given that AB is the height of the man, the angle of elevation is 60°, the distance from the man to the tower BC is 36 m and the height of the tower is 64 m.

To find:

The height of the man AB.

Solution:

In ΔADE

$tan 60=\frac{ED}{AD}$

⇒$\sqrt{3}=\frac{ED}{36}$

⇒$36\sqrt{3}=ED$

Now,

DC=CE-CD

⇒$CD=64-36\sqrt{3}$

⇒$CD=4(16-9\sqrt{3})$

Now, we know that AB=CD

Therefore, The height of the man is $4(16-9\sqrt{3})$