Question

the angle of elevation of the top of a tower from a point on the ground which is 15cm away from the foot of the tower is 60° the height pf the tower​

Answer 1

$\huge{\orange{\underline{\red{\mathbb{ANSWER:-}}}}}$

HEIGHT $\huge\bold\red\rightarrow$ 15√3.

$\large{\orange{\underline{\red{\mathbb{STEP \: BY \: STEP \: EXPLANATION:- }}}}}$

Let AB be the height of the tower.

Now,

In△ABC,

length of BC Given 15m.

and also Given angle of elevation 60° .

$\huge\bold\red\rightarrow$ Tan 60° = AB/CB

$\huge\bold\red\rightarrow$ √3 = H/15

$\huge\bold\red\rightarrow$ 15 √3 = H

$\huge\bold\red\rightarrow$ H = 15√3

Hence,

$\huge\bold\blue\rightarrow$ The height of the tower is 15 √3m

$\huge{\orange{\underline{\red{\mathbb{THANKS.}}}}}$