Question

The angle of elevation of a tower from a dis-
tance 100 m from its foot is 30°. Height of the
tower is:
(a) 100 V3 m
(b) 5 m
(C) 50 v3 m
m​

Answer 1

Answer:

100 √ 3 / 3 m .

Step-by-step explanation:

Given :

Angle ( Ф )  =  30⁰

Distance = 100 m

We have to find height of tower .

Let say height be h m.

Now we know

tan Ф  = perpendicular / base  = P / B

putting the values here we get

tan 30 = h / 100

tan 30 = 1 / √ 3

1 / √ 3 = h / 100

h = 100 / √ 3 m

Now rationalize the denominator :

100 × √ 3 / 3

100 √ 3 / 3 m

Thus the height of tower is 100 √ 3  / 3 m .

Answer 2

Given :-

Distance from foot of tower is 100m.

$\theta = 30^{\circ}$

To find :-

The height of tower.

Explanation :-

Let the height of tower be H metre.

• See in above attachment.

→$In \triangle{ABC}, \angle{B}=90^{\circ}$

→$Tan\theta = \dfrac{P}{B}$

→$Tan\theta = \dfrac{AB}{BC}$

→$Tan30^{\circ} = \dfrac{H}{100}$

→$\dfrac{1}{\sqrt{3}}= \dfrac{H}{100}$

→$\sqrt{3}H = 100$

→$H = \dfrac{100}{\sqrt{3} }m$

hence,

The height of tower is $\dfrac{100}{\sqrt{3} }m$