Question

15. A tower stands vertically on the ground. From a point on the ground, which is 15m away
from the foot of the tower, the angic of elevation of the fop of the tower is found to be 60
Find the height of the tower.​

Answer 1

Answer: taken triangle triangle ABD right angled at  B

               AC is the hyp

n 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° =AB/CB

root 3= H/15

15root 3=H

H=15root 3

Hence,

The height of the tower is  

hope it helps

tq,./

Step-by-step explanation:

Answer 2

Step-by-step explanation:

GIVEN :-

$\sf \: Distance \: of \: the \: point \: from \: the \: foot \: of \: tower \\ \sf =15 \: m$

$\sf \: Angle \: of \: elevation ( \theta) \: = 60 \degree$

TO FIND :-

$\sf \: Height \: of \: tower \: (h) \:$

SOLUTION :-

$\sf \: Let \: height \: of \: tower \: AB \: is \: ( h) \: m \\ \sf \: now \: \: in \: \triangle \: ABC \\ \\ \rightarrow \sf \: \tan(60) = \frac{AB}{BC} \\ \\ \rightarrow \sf \: \sqrt{3} = \frac{h}{15} \: \: \: \: \: \: \: \: \: \: \: \\ \\ \rightarrow h = 15 \sqrt{3 \:} m \: \: \: \:$

$\sf \: Hence \green{ \boxed{ \: Height \: of \: tower \: is \: \: 15 \sqrt{3} \: m}}$