Question

An aeroplane is at a vertical height of 675 m from the bank of a river. The angle of elevation of the aeroplane from a point just on the opposite bank of that river is found to be 60${}^o$. The width of the river is _____ m.

• A. 225$\sqrt{3}$
• B. 215$\sqrt{3}$
• C. 205$\sqrt{3}$
• D. None of these

Answer 1

The correct option is A 225$\sqrt{3}$

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

Let $A$ be the aeroplane at a vertical height of $675m$ above the bank of a river the angle of elevation of the aeroplane from a point just on the opposite bank is found to be ${60}^o$. Let that point be $C$.

We know that, $\tan \theta =\frac{\text{Opposite Side}}{\text{Adjacent Side}}$

Hence, $\tan {60}^o=\frac{AB}{BC}$

$BC=\frac{AB}{\tan {60}^o}$

$=\frac{675}{\sqrt{3}}[\because \tan 60°=\sqrt{3}]$

$=\frac{675\times \sqrt{3}}{\sqrt{3}\times \sqrt{3}}$

$=225\sqrt{3}$

Thus, width of the river is $225\sqrt{3}m$.