Question

A person observed the angle of elevation of the top of a tower as ${30}^{\circ }$,he walks 50m towards the foot of the tower along level ground and found the angle of elevation of the top of tower ${60}^{\circ }$,find the height of the tower.

Answer 1

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$\tan {60}^{\circ }=\frac{AB}{BC}=\sqrt{3}$

$\Rightarrow BC=\frac{AB}{\sqrt{3}}$

Now $\Delta B=DC+BC=50+\frac{AB}{\sqrt{3}}$

$tan{30}^{\circ }=\frac{1}{\sqrt{3}}=\frac{AB}{\Delta C}=\frac{AB}{50+\frac{AB}{\sqrt{3}}}$

$\Rightarrow 50+\frac{AB}{\sqrt{3}}=\sqrt{3}AB$

$\Rightarrow 50\sqrt{3}+AB=3AB$

$Ab=25\sqrt{3}=$ Height of towar