Question

The angle of elevation of an airplane from a point on the ground is ${45}^{\circ }$ After the flight of $15$ sec the elevation changes to ${30}^{\circ }$ If the airplane is flying at a height of $3000$ meters find the speed of the airplane.

Answer 1

In $△ABC,$

$\Rightarrow \tan 45°=\frac{AC}{BC}$

$\Rightarrow 1=\frac{3000}{BC}$

$\Rightarrow BC=3000m$

Again, in $△EBD,$

$\Rightarrow \tan 30°=\frac{ED}{BD}=\frac{ED}{BC+CD}$

$\Rightarrow \frac{1}{\sqrt{3}}=\frac{3000}{3000+CD}$

$\Rightarrow 3000+CD=3000\sqrt{3}$

$\Rightarrow CD=3000(\sqrt{3}-1)$

Now, CD distance is covered in $15sec,$ so that

$\Rightarrow$ Speed $=\frac{3000(\sqrt{3}-1)}{15}=200(\sqrt{3}-1)m/sec$

Hence, the answer is $200(\sqrt{3}-1)m/sec.$