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.   

Answer 1

Given:
Distance between points A and B = 500 km
B from A is 30˚ east of north.
Speed of wind due north, v_w = 20 m/s
Airspeed of the plane, v_a = 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.