A bird is sitting on the top of a tree, which is 80m high. The angle of elevation of the bird, from a point on the ground is $45^{\circ}$. The bird flies away from the point of observation horizontally and remains at a constant height. After 2 seconds, the angle of elevation of the bird from the point of observation becomes $30^{\circ}$. Find the speed of the bird.
$B_1$: Initial position of the bird
$B_2$: Final position of the bird
$B_1B_2$: Distance the bird flies in 2 seconds
A: Point of observation
From the figure, $OB_1=PB_2 = 80m$
In triangle $OB_1A$,
$\dfrac{OB_1}{OA}=\tan 45^{\circ} = 1$ $\implies$ $OA=OB_1=80$
In triangle $PB_2A$,
$\dfrac{PB_2}{PA}=\tan 30^{\circ} = \dfrac{1}{\sqrt{3}}$ $\implies$ $PA=\sqrt{3}PB_2=80\sqrt{3}$
Now, $PO=PA-OA=80\sqrt{3}-80=58.56$
$\therefore \;$ Distance the bird flies $=B_1B_2=OP=58.56m$
$\therefore \;$ Speed of the bird $= \dfrac{\text{Distance covered by the bird}}{\text{Time taken}} = \dfrac{58.56}{2}=29.28m/s$