Question

A tower stands vertically on the ground from a point on the ground which is 15 M away from the foot of the tower and the angle of elevation of the top of the tower is found to be 60 degree find the height of the tower

Answer 1

$\huge{Hey\:Mate!}$

here is your answer !!

in the questions we find the height of the tower.

Let AB be the height of the tower.

Now,

In△ABC,

length of BC Given 15m.

and also Given angle of elevation 60° .

Tan 60°$= \frac{AB}{CB}$

$\sqrt{3} = \frac{H}{15}$

$15 \sqrt{3} = H$

$H = 15 \sqrt{3}$

Hence,

The height of the tower is $15 \sqrt{3m}$

hope it's help you!!

$\huge{Be \:Brainly!}$

Answer 2

$\underline{\underline{\mathfrak{\large{Solution:}}}}$

Let $XY$ be the height of the tower.

Now,

In △$XYZ,$

Tan 60° $= \frac{XY}{ZY}$

$\sqrt{3} = \frac{H}{15}$

$15 \sqrt{3} = H$

$H = 15 \sqrt{3}$

Hence,

The height of the tower is $15 \sqrt{3}m.$

#Be Brainly❤️