# The angle of elevation of the top of a tower $30\ m$ high from the foot of another tower in the same plane is $60^o$ and the angle of elevation of the top of the second tower from the foot of the first tower is $30^o$. then find the distance between the two towers.

**Given:** The angle of elevation of the top of a tower $30\ m$ high from the foot of another tower in the same plane is $60^o$ and the angle of elevation of the top of the second tower from the foot of the first tower is $30^o$.

**To do:** To find the distance between the two towers.

**Solution:**

Let the distance between the two towers$=AB=x\ m$ and height of the other tower$=PA=h\ m$

Given that height of the tower$=QB=30\ m$ and $\angle QAB=60^o,\ \angle PBA=30^o$

Now, in $\vartriangle QAB$,

$\ tan60^o=\frac{QB}{AB}=\frac{30}{x}$

$\Rightarrow \sqrt{3}=\frac{30}{x}$

$\Rightarrow x=\frac{30}{\sqrt{3}}.\frac{\sqrt{3}}{\sqrt{3}}$

$\Rightarrow x=\frac{30\sqrt{3}}{3}=10\sqrt{3}\ m$

Thus, the distance between the towers is $10\sqrt{3}\ m$.

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