Question

From a point on the ground, the angle of elevation of the top of a tower is observed to be ${60}^{\circ }$. From a point 40$m$ vertically above the first point of observation, the angle of elevation of the top of the tower is ${30}^{\circ }$. Find the height of the tower and its horizontal distance from the point of observation.

Answer 1

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To find $\to$ Height of tower (AB)

Let AB $=h$ m

In $△ABC$, by Trigonometry

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

$\sqrt{3}=\frac{h}{BC}\Rightarrow BC=\left(\frac{h}{\sqrt{3}}\right)m$

Since BCDE is a ||gm, so $BC=DE$ and CD = BE

$\therefore DE=\left(\frac{h}{\sqrt{3}}\right)m;BE=40m$

Now in $△ADE$

$tan{30}^{\circ }=\frac{AE}{DE}$

$\frac{1}{\sqrt{3}}=\frac{h-40}{\frac{h}{\sqrt{3}}}$

$h=3h-120$

$2h=120m$

$h=60m$

Height of tower = 60 m.

To find $\to$ Horizontal Distance from point of observation (BC)

$BC=\frac{h}{\sqrt{3}}=\frac{60}{\sqrt{3}}=20\sqrt{3}m$

$BC=20\sqrt{3}m=34.60m$