Question

A boy standing on a horizontal plane finds a bird flying at a distance of $100m$ from him at an angle of elevation of ${30}^0$. A girl standing on the roof of $20m$ high building. If the angle of elevation of the same bird to be ${45}^0$. Both the boy and the girl are opposite sides of the bird. Find the distance of the bird from the girl.

Answer 1

In the following figure let $A,E$ and $C$ are the position of bird, girl and boy respectively.

Now

$\angle ACB={30}^0,\angle AED={45}^0$

$AC=100m$ and $EF=20m$

From right angled $\Delta ABC$,

$\sin {30}^0=\frac{AB}{AC}$

$\frac{1}{2}=\frac{AB}{100}$

$AB=\frac{100}{2}=50m$

$AB=AD+BD$

$50=AD+20m$

$AD=50–20=30m$

From right angled $\Delta ADE$,

$\sin {45}^0=\frac{AD}{AE}$

$\frac{1}{\sqrt{2}}=\frac{30}{AE}$

$AE=30\sqrt{2}=30\times 1.41$

$=42.3m$

Hence, distance of bird from girl = $42.3m$