Question

A pole $5m$ high is fixed on the top of a tower. The angle of elevation of the top of the poles observed from a point $A$ on the ground is ${60}^{\circ }$ and the angle of depression of point $A$ from the tower is ${45}^{\circ }$. Find the height of the tower $\sqrt{3}=1.732$.

Answer 1

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$AC=100-h$

$tan{45}^{\circ }=\frac{100-h}{x}$
$\Rightarrow 1=\frac{100-h}{x}$

$\Rightarrow x=100-h$
$\Rightarrow h=100-x$.......(1)

Now $tan{60}^{\circ }=\frac{100}{x}=\sqrt{3}$
$\Rightarrow \sqrt{3}=\frac{100}{x}$

$\Rightarrow x=\frac{100\sqrt{3}}{3}$
Substitute the value of x in equation$\left(1\right)$, we get

$Nowh=100-\frac{100\sqrt{3}}{3}$

$=100(1-\frac{\sqrt{3}}{3})$
$=\frac{100}{3}(3-\sqrt{3})$