Discussion
An aeroplane when 900 m high passes vertically above another aeroplane at an instant when their angles of elevation at same observing point are 60° and 45° respectively. Approximately, how many meters
Answer: D
Both aeroplanes lie on the same vertical line, so a single horizontal distance \( d \) serves both sightings.
Upper plane, 900 m high, elevation \( 60^{\circ} \):
\[ d = \frac{900}{\tan 60^{\circ}} = \frac{900}{\sqrt{3}} = 300\sqrt{3} = 519.6 \text{ m} \]
Lower plane, elevation \( 45^{\circ} \):
\[ h = d \tan 45^{\circ} = 519.6 \text{ m} \]
\[ 900 - 519.6 = 380.4 \approx 381 \text{ m} \]
The larger elevation must belong to the higher aircraft, which is what pairs \( 60^{\circ} \) with 900 m.
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