An aeroplane has to go from a point $A$ to another point B, $500 k m$ away due $30 °$ east of north. Wind is blowing due north at a speed of $20 m / s$ . The steering-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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$150 ° - θ [or \sin^{- 1} (\frac{1}{15})]$ , east of the line AB

$150 ° - θ [or \sin^{- 1} (\frac{1}{15})]$ , west of the line AB

$150 ° - θ [or \sin^{- 1} (\frac{1}{15})]$ , north of the line AB

None of the above

answer is A.

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Detailed Solution

Applying Sine law in $∆ A B C$ , we have

$\frac{150}{\sin 30 °} = \frac{y}{\sin θ} = \frac{20}{\sin (150 ° - θ)}$

From and third, we have,

$\sin (150 ° - θ) = \frac{20 \sin 30 °}{150} = \frac{1}{15} 150 ° - θ = \sin^{- 1} (\frac{1}{15})$

Hence, we can see that direction of 150 ms^-1 is $150 ° - θ [or \sin^{- 1} (\frac{1}{15})]$ , east of the line AB.

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