Question

The angles of elevation of an artificial earth satellite is measured from two earth stations, situated on the same side of the satellite, are found to be ${30}^{\circ }$ and ${60}^{\circ }$. The two earth stations and the satellite are in the same vertical plane. If the distance between the earth stations is $4000$ km, find the distance between the satellite and earth. $(\sqrt{3}=1.732)$

• A. $3446$ km
• B. $3464$ km
• C. $3488$ km
• D. $3400$ km

Answer 1

The correct option is B $3464$ km

Consider A and B be two earth stations

Let AB $=4000$ km

Let BC $=y$ km and CD $=x$ km, where CD is the height of the satellite from the earth

From $△ACD$

$\tan {30}^{\circ }=\frac{x}{4000+y}$

$⟹\frac{x}{4000+y}=\frac{1}{\sqrt{3}}$

$⟹\sqrt{3}x=4000+y$ ........(1)

From $△BCD$

$\tan {60}^{\circ }=\frac{x}{y}$

$⟹\frac{x}{y}=\sqrt{3}$

$⟹y=\frac{x}{\sqrt{3}}$ ........(2)

From (1) and (2)

$\sqrt{3}x=4000+\frac{x}{\sqrt{3}}$

$⟹3x=4000\sqrt{3}+x$

$⟹2x=4000\sqrt{3}$

$⟹x=2000\sqrt{3}\text{km}$

$⟹x=2000\times 1.732=3464\text{km}$ $[\because \sqrt{3}=1.732]$

Hence, the distance of satellite from the earth is $3464\text{km}$