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The angle of elevation of an aeroplane from a point on the ground is 60°. After 15 seconds flight, the elevation changes to 30°. If the aeroplane is flying at a height of 1500 √3 m, find the speed of the plane
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- 300 m/sec
- 200 m/sec
- 100 m/sec
- 150 m/sec
Correct Option: B
P & Q are the positions of the plane.
∠PAB = 60° ; ∠QAB = 30°
PB = 1500 √3 metre
In ∆ ABP,
| tan 60° = | AB |
| ⇒ √3 = | AB |
⇒ AB = 1500 metre
In ∆ ACQ,
| tan 30° = | AC |
| = | = | |||
| √3 | AC |
= AC = 1500 × 3 = 4500 metre
PQ = BC = AC – AB
= 4500 – 1500 = 3000 metre
⇒ 3000 m travelled in 15 sec.
| ∴ Speed of plane = | 15 |
= 200 metre/second
