Question

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 $2$ secs, the angle of elevation of the bird from the point of observation becomes ${30}^{\circ }.$ Find the speed of the flying bird.

Answer 1

In the above figure bird initially, at point, $A$ flies to point $B$ in $2$ secs.

Let the speed of the bird be $v$ then $CD$ will be $2v$ as the distance is equal to the product of speed and time.

In $△ABC,$

$\begin{array}{cc}\tan {45}^{\circ }& =\frac{AC}{BC}\\ 1& =\frac{80}{BC}\\ BC& =80m\end{array}$

Now, in $△EDB,$

$\begin{array}{cc}\tan {30}^{\circ }& =\frac{ED}{BD}\\ \frac{1}{\sqrt{3}}& =\frac{80}{BC+CD}\\ \frac{1}{\sqrt{3}}& =\frac{80}{80+2v}\\ 80+2v& =80\sqrt{3}\\ 2v& =80\left(\sqrt{3}-1\right)\\ v& =40\left(\sqrt{3}-1\right)m/s\end{array}$

Hence, speed of the flying bird is equal to $40(\sqrt{3}-1)m/s.$