Question

An aeroplane when flying at a height of $3000m$ from the ground passes vertically above another aeroplane at an instant when the angles of elevation of the two planes from the same point on the ground are ${60}^0$ and ${45}^0$ respectively. Find first aeroplane is how much high with second aeroplane.

Answer 1

Let height of first aeroplane is $BD$ which is flying at $3000m$ height and height of second aeroplane $BC=hm$. Let angles of elevation from point $A$ are ${60}^0$ and ${45}^0$ respectively.

So, $\angle BAD={60}^0$ and $\angle CAB={45}^0$ and Let $AB=xm$

From right angled $\Delta ABD$,

$\tan {60}^0=\frac{BD}{AB}$

$\sqrt{3}=\frac{3000}{x}$

From right angled $\Delta ABC$,

$\tan {45}^0=\frac{BC}{AB}$

$1=\frac{h}{x}$

$1=\frac{h}{1000\sqrt{3}}$

$h=1000\sqrt{3}=1000\ast 1.732$

$=1732m$

Hence, height of second plane from first plane

$CD=BD–BC=3000–1732=1268m$