Question

The angles of elevation of an artificial satellite as measured from two earth stations are ${30}^{\circ }$ and ${60}^{\circ }$. If the distance between the earth stations is 4000 km, the height of the satellite is

• A. 2800 km
• B. 2000 km
• C. 3152 km
• D. 3465 km

Answer 1

The correct option is D 3465 km


Height of satellite = AB
In $\Delta ABD$,
$tan{30}^{\circ }=\frac{AB}{4000+CB}$
$\Rightarrow \frac{1}{\sqrt{3}}=\frac{AB}{4000+CB}$
$\Rightarrow CB=(\sqrt{3}AB-4000)km\dots \dots \left(1\right)$
In $\Delta ACB$,
$tan{60}^{\circ }=\frac{AB}{CB}$
$\Rightarrow \sqrt{3}=\frac{AB}{CB}$
$\Rightarrow CB=\left(\frac{AB}{\sqrt{3}}\right)km\dots \dots \left(2\right)$
From (1) and (2), we get
$\sqrt{3}AB-4000=\frac{AB}{\sqrt{3}}$
$\Rightarrow 3AB-4000\sqrt{3}=AB$
$\Rightarrow 2AB=4000\sqrt{3}$
$\Rightarrow 2AB=6428.2$
$\Rightarrow AB=3464.10km=3465km$