Question

The angle of elevation of a tower from a distance 100 m from its foot is 30°. Height of the tower is:

Answer 1

1. Answer: height of the tower = 57.6 metres  Step-by-step explanation: Angle of elevation = 30° Distance = 100 metres  Let the height of the tower be HH’ where H is the foot of the tower and H’ is the topmost point of the tower  The distance from the point of elevation to the foot of the power be AH, where A is the point from which the angle of elevation is taken   Tan30° = 1/√3 = AH / HH’ => HH’ / √3 = AH => 100 / √3 = AH => AH = 100√3 / 3 = (100 * 1.73) / 3 = 173/3 = 57.6 metres   Therefore height of the tower = 57.6 metres      Please brainlist my answer, if helpful!

Answer 2

$\textbf{\underline{\underline{According\:to\:the\:Question}}}$

Angle from its foot = 30

Distance = 100 m

Now

Height of tower

Assumption

Height = h

Using trigonometric ratios :-

$\tt{\rightarrow tan\theta=\dfrac{Perpendicular}{Base}}$

Substituting the values

$\tt{\rightarrow tan30=\dfrac{h}{100}}$

$\tt{\rightarrow\dfrac{1}{\sqrt{3}}=\dfrac{h}{100}}$

$\tt{\rightarrow h=\dfrac{100}{\sqrt{3}}}$

Here,

We have to rationalize :-

$\tt{\rightarrow 100\times\dfrac{\sqrt{3}}{3}}$

$\tt{\rightarrow 100\times\dfrac{\sqrt{3}}{3}}$

Hence,

Height of tower :-

$\tt{\rightarrow 100\times\dfrac{\sqrt{3}}{3}m}$

$\tt{\rightarrow\dfrac{100\times 1.73}{3}}$

$\tt{\rightarrow\dfrac{173}{3}}$

= 57.6 metres