Heights and Distances

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

$X$: Position of bird

$B$: Position of boy

$OH$: Building on whose roof the girl is standing

Height of the building $= OH = 20 \; m$

Distance of the bird from the boy $= BX = 100 \; m$

Distance of the bird from the girl $= HX$

In $\triangle BPX$, $\;$ $\dfrac{PX}{BX} = \sin 30^\circ = \dfrac{1}{2}$

$\implies$ $PX = \dfrac{1}{2} \times BX = \dfrac{1}{2} \times 100 = 50 \; m$

Draw $HQ \perp PX$

Then, $PQ = OH = 20 \; m$

$\therefore \;$ $QX = PX - PQ = 50 - 20 = 30 \; m$

In $\triangle HQX$, $\;$ $\dfrac{QX}{HX} = \sin 45^\circ = \dfrac{1}{\sqrt{2}}$

$\implies$ $HX = \sqrt{2} \times QX = \sqrt{2} \times 30 = 42.42$

$\therefore \;$ Distance of the bird from the girl $= 42.42 \; m$