1. Set up the directions

Let the river flow along the positive $x$-axis with speed $v$.

The swimmer’s speed relative to water is also $v$.

The swimmer wants the resultant velocity (with respect to ground) to be along line $AB$.

Given: line $AB$ makes an angle of $30^\circ$ with the flow of river.

So the resultant velocity must make angle $30^\circ$ with the $x$-axis.

2. Represent the velocities

• Velocity of river: $\vec v_r = v\hat i$

• Velocity of swimmer relative to water: magnitude $v$, making some angle such that resultant goes along $AB$.

Let the resultant velocity be $\vec v_{\text{res}} = \vec v_s + \vec v_r$

and it must make angle $30^\circ$ with the flow direction.

3. Use the fact that magnitudes are equal

Since both speeds are equal to $v$, and one vector is the river velocity and the other is swimmer’s velocity relative to water, we need their sum to point along $AB$ at $30^\circ$.

Take the swimmer’s velocity to make angle $\phi$ with the flow direction. Then $\vec v_s = v(\cos\phi\,\hat i + \sin\phi\,\hat j)$

Hence, $\vec v_{\text{res}} = v(1+\cos\phi)\hat i + v\sin\phi\hat j$

Since resultant makes angle $30^\circ$ with the flow direction, $\tan 30^\circ = \frac{v\sin\phi}{v(1+\cos\phi)}$

So, $\frac{1}{\sqrt3} = \frac{\sin\phi}{1+\cos\phi}$

But $\frac{\sin\phi}{1+\cos\phi} = \tan\frac{\phi}{2}$

Therefore, $\tan\frac{\phi}{2} = \frac{1}{\sqrt3}$

Thus, $\frac{\phi}{2} = 30^\circ \quad \Rightarrow \quad \phi = 60^\circ$

4. Find the required angle with line $AB$

• Swimmer’s velocity makes angle $60^\circ$ with the river flow.
• Line $AB$ makes angle $30^\circ$ with the river flow.

Therefore, the angle between swimmer’s direction and line $AB$ is $\theta = 60^\circ - 30^\circ = 30^\circ$

5. Final answer

$\boxed{30^\circ}$