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Question
An aeroplane has to go from a point A to another point B, 500 km away due 30° east of north. A wind is blowing due north at a speed of 20 m/s. The air-speed of the plane is 150 m/s. Find the direction in which the pilot should head the plane to reach the point B.
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Solution
Given:
Distance between points A and B = 500 km
B from A is 30˚ east of north.
Speed of wind due north, vw = 20 m/s
Airspeed of the plane, va = 150 m/s
Let \[\vec{R}\] be the resultant direction of the plane to reach point B.
Using the sine formula in ∆ACB, we get:
\[\frac{20}{\sin A} = \frac{150}{\sin30^\circ}\]
\[ \Rightarrow \sin A = \frac{20}{150}\sin30^\circ\]
\[ = \frac{20}{150} \times \frac{1}{2} = \frac{1}{15}\]
\[ \Rightarrow A = \sin^{- 1} \left( \frac{1}{15} \right)\]
Direction of the aeroplane is \[\sin^{- 1} \left( \frac{1}{15} \right)\] east of line AB.
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