Question

A tower stand vertically on the ground. From a point on the ground which is 18m away from the foot of tower, the angle of elevalion of the top of
tower is found to be 30°. Find height of tower.​

Answer 1

=> Angle of elevation of the top of tower = 30°

=> Distance between the point on ground and foot of tower = 18 m

Explanation of diagram ( refer attachment for diagram ) :-

• AB = tower

• C is the point on ground

• B is foot of tower

• CB = 18 m

We know that :-

$\implies \sf tan \theta = \dfrac{perpendicular}{base }$

$\implies \sf tan 30\degree = \dfrac{1}{ \sqrt{3} }$

Let height of tower be h

Now, according to the question :-

$\sf \implies tan 30 \degree = \dfrac{h}{18}$

$\sf \implies \dfrac{1}{ \sqrt{3} } = \dfrac{h}{18}$

$\sf \implies \dfrac{1}{ \sqrt{3} } \times 18 = {h}$

$\sf \implies \dfrac{18}{ \sqrt{3} } = {h}$

$\sf \implies \dfrac{18}{ \sqrt{3} } \times \dfrac{ \sqrt{3} }{ \sqrt{3} } = {h}$

$\sf \implies \dfrac{18 \sqrt{3}}{ 3} = {h}$

$\sf \implies \dfrac{6\sqrt{3}}{ 1} = {h}$

$\sf \implies h = 6\sqrt{3} \: m$

Answer :

• Height of tower = 6√3 m

Answer 2

$\huge\underline{\bf \orange{AnSweR :}}$

=> Angle of elevation of the top of tower = 30°

=> Distance between the point on ground and foot of tower = 18 m

AB = tower

C is the point on ground

B is foot of tower

CB = 18 m

We know that :-

Base/Perpendicular=> tan30°= 3//1

Let height of tower be h

Now, according to the question :-

=> tan30°= 18/h

=> 3/1 = 18h

=> 3/1 ×18=h

=> 18³ =h

=> 3/ 18×3³=h

=> 3/18³ =h

=> 1/6×3 =h

=> h=6×3 m

Height of tower = 6√3m Ans.