Question

An airplane has to go from a point A to Point B, $500km$ away due $30°$ East of the north. A wind is blowing due north at a speed of $20m/s$. The air speed of the plane is $150m/s$. Find the direction in which the pilot should head the plane to reach point B.

Answer 1

Step 1: Given

Distance between points A and B: $D=500km$

Angle in north-east direction: $\theta =30°$

The velocity of the wind: $V_w=20m/s$

The velocity of the plane: $V_a=150m/s$

Step 2: Formula Used

$$\begin{gathered} In∆ABC, \\ \frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c} \end{gathered}$$

Where, $a$ is length of $BC$, $b$ is length of $AC$ and $c$ is length of $AB$.

Step 3: Find the direction of the plane

Calculate the value of A by using the sine formula on $∆ACD$.

$$\begin{gathered} \frac{\sin A}{20}=\frac{\sin 30}{150} \\ \Rightarrow \sin A=\frac{20}{150}\sin 30 \\ \Rightarrow \sin A=\frac{20}{150}\times \frac{1}{2} \\ \Rightarrow A={\sin }^{-1}\left(\frac{1}{15}\right) \end{gathered}$$

Hence, the direction must be ${\sin }^{-1}\left(\frac{1}{15}\right)$ towards east of AB.