The angle of elevation of a tower from a distance 100 m from its foot

1.	The angle of elevation of a tower from a distance 100 m from its foot is 30°. Height of the tower is:
Answer» $\textbf{\underline{\underline{According\:to\:the\:Question}}}$ 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