Question

Rajender observes a person standing on the ground from a helicopter at an angle of depression ${45}^o$, If the helicopter flies at a height of $500meters$ from the ground, what is the distance of the person from Rajender?

Answer 1

Based on the given information, we can draw the figure shown above.

$In\Delta ABC,\tan 45°=\frac{BC}{AC}$

$1=\frac{BC}{AC}[\because \tan {45}^o=1]$

$AC=500m$

Since, triangle ABC is right angled triangle,

$A{B}^2=A{C}^2+B{C}^2$

$\Rightarrow A{B}^2={500}^2+{500}^2$

$\therefore AB=500\sqrt{2}m$

Distance between them is $500\sqrt{2}m$