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 $2$ seconds, the angle of elevation of the bird from the point of observation becomes ${30}^{\circ }.$ Find the speed of flying of the bird (Take $\sqrt{3}=1.73$ ).

Answer 1

$\tan {45}^o=\frac{AC}{BC}$

$1=\frac{80}{BC}$

$BC=80$cm

$\tan {30}^o=\frac{DE}{BC+CE}$

$\frac{1}{\sqrt{3}}=\frac{80}{80+x}$

$80+x=80\sqrt{3}$

$x=80\sqrt{3}-80$

$x=80(\sqrt{3}-1)m$

The bird travelled $80(\sqrt{3}-1)m$

Speed of bird $=\frac{Distance}{Time}$

$=\frac{40(\sqrt{3}-1)}{2}$

$=40(\sqrt{3}-1)m/s$

$\therefore$ Speed $=40(\sqrt{3}-1)m/s$,