Discussion
An aeroplane flying at a height of 4000 m passes vertically above another aeroplane at the instant when the angles of elevation of the two planes from the same point on the ground are \( 60^{\circ} \) and \( 45^{\circ} \) respectively. Find the vertical distance between the two planes. (Take \( \sqrt{3} = 1.732 \))
Answer: C
Both planes lie on the same vertical line, so one horizontal distance \( d \) serves both sightings. The higher plane takes the steeper angle.
\[ \tan 60^{\circ} = \frac{4000}{d} \quad\Longrightarrow\quad d = \frac{4000}{1.732} = 2309.47 \text{ m} \]
\[ h = d\tan 45^{\circ} = 2309.47 \text{ m} \]
\[ \text{vertical gap} = 4000 - 2309.47 = 1690.53 \text{ m} \]
At \( 45^{\circ} \) the lower plane's height simply equals the horizontal distance, which is why only one division is really needed.
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