An aeroplane is required to travel from a point A to another point...

Let $\vec{v}_{p}$ be the velocity of the plane with respect to the ground, $\vec{v}_{w}$ be the velocity of the wind, and $\vec{v}_{pw}$ be the velocity of the plane with respect to the wind.

We are given that $\vec{v}_{p} = \vec{v}_{pw} + \vec{v}_{w}$.

Let the north direction be along the y-axis ($\hat{j}$) and the east direction be along the x-axis ($\hat{i}$).

The wind velocity is $\vec{v}_{w} = 20 \hat{j}$.

Let the plane head at an angle $\alpha$ east of north. The velocity of the plane with respect to the wind is:

$\vec{v}_{pw} = 150 \sin \alpha \hat{i} + 150 \cos \alpha \hat{j}$

The velocity of the plane with respect to the ground becomes:

$\vec{v}_{p} = 150 \sin \alpha \hat{i} + (150 \cos \alpha + 20) \hat{j}$

For the plane to travel along the line $AB$, which is $30^\circ$ east of north, the resultant velocity $\vec{v}_{p}$ must also be directed $30^\circ$ east of north. Thus, the ratio of its x-component to its y-component must be $\tan 30^\circ$:

$\dfrac{150 \sin \alpha}{150 \cos \alpha + 20} = \tan 30^\circ = \dfrac{1}{\sqrt{3}}$

Rearranging the equation:

$150 \sqrt{3} \sin \alpha = 150 \cos \alpha + 20$

$150 (\sqrt{3} \sin \alpha - \cos \alpha) = 20$

$\sqrt{3} \sin \alpha - \cos \alpha = \dfrac{20}{150} = \dfrac{2}{15}$

Dividing both sides by $2$:

$\dfrac{\sqrt{3}}{2} \sin \alpha - \dfrac{1}{2} \cos \alpha = \dfrac{1}{15}$

Using the trigonometric identity $\sin A \cos B - \cos A \sin B = \sin(A - B)$:

$\sin \alpha \cos 30^\circ - \cos \alpha \sin 30^\circ = \dfrac{1}{15}$

$\sin(\alpha - 30^\circ) = \dfrac{1}{15}$

$\alpha - 30^\circ = \sin^{-1}$\left(\dfrac{1}{15}\right)

Since $\alpha$ is the angle east of north and the line $AB$ is $30^\circ$ east of north, the difference $\alpha - 30^\circ$ represents the angle east of the line $AB$.

Therefore, the pilot should head $\sin^{-1}$\left(\dfrac{1}{15}\right)$east of the line$AB$.<br><br>Answer:$\sin^{-1} $\left(\dfrac{1}{15}\right) east of the line$AB