Question

A bird was flying in a line parallel to the ground at a height of $2000$ metres. Tom, standing in the middle of the field, he observed the bird making an angle of elevation ${30}^{\circ }$. After $3$ minutes, he again observed it with an angle of elevation of ${45}^{\circ }$, the speed of the bird in kilometers per hour $(\sqrt{3}=1.732)$ is:

Answer 1

Height of the bird from the ground $=AB=CD=2000$ m
Initially angle of elevation = ${30}^{\circ }$
After $3$ minutes, angle of elevation = ${45}^{\circ }$

Let the distance when angle is ${30}^{\circ }$ be $d_1$ and when angle is ${45}^{\circ }$ be $d_2$
Now, from $\Delta ABE$

$\tan {30}^o=\frac{AB}{BE}$

$\Rightarrow \frac{1}{\sqrt{3}}=\frac{2000}{d_1}$
$\Rightarrow d_1=2000\sqrt{3}$

And, from $\Delta CDE$

$\tan {45}^o=\frac{CD}{DE}$

$\Rightarrow 1=\frac{2000}{d_2}$

$\Rightarrow d_2=2000$

Now, distance travelled in 3 minutes, $=d_1-d_2=(2000\sqrt{3}-2000)m$

$=(2000\times 1.732-2000)m$

$=1464m$
Now, $1464m=1.464km$

Thus, speed = $\frac{distance}{time}$
speed = $\frac{1.464}{\frac{3}{60}}km/hr$
speed = $29.28km/hr$