A swimmer wants to cross a river from point $$A$$ to point $$B$$. Line AB makes an angle of 30° with the flow of the river. The magnitude of the velocity of the swimmer is the same as that of the river. The angle $$\theta$$ with the line AB should be _________ °, so that the swimmer reaches point $$B$$.

We need to determine the angle $$\theta$$ relative to the line $$AB$$ that a swimmer must maintain so that their net resultant path aligns perfectly along $$AB$$.

1. Identify the Motion Vectors

From the problem details and the accompanying vector diagram page:

• The line $$AB$$ is the desired path of the swimmer and makes an angle of $$30^\circ$$ with the river flow direction.
• Let the velocity of the river be $$\vec{v}_r$$ directed horizontally to the right.
• The magnitude of the swimmer's velocity relative to the water ($$v_s$$) is equal to the magnitude of the river's velocity ($$v_r$$). Therefore, we can set $$v_s = v_r = v$$.
• The swimmer heads upstream at an angle $$\theta$$ with respect to the line $$AB$$ to counteract the river's push.

2. Analyze Components Perpendicular to the Path AB

For the swimmer to travel directly from point $$A$$ to point $$B$$, their net velocity component perpendicular to the line $$AB$$ must be exactly zero. This means the perpendicular component of the swimmer's velocity relative to the water must perfectly balance the perpendicular component of the river's velocity:

• River Velocity Component Perpendicular to AB:
  The river velocity vector makes an angle of $$30^\circ$$ with the line $$AB$$. Its component perpendicular to the line $$AB$$ is:

  $$v_{\text{river, }\perp} = v_r \sin 30^\circ$$

• Swimmer Velocity Component Perpendicular to AB:
  The swimmer's velocity vector makes an angle $$\theta$$ with the line $$AB$$. Its component perpendicular to the line $$AB$$ is:

  $$v_{\text{swimmer, }\perp} = v_s \sin \theta$$

3. Equate and Calculate

Equating both perpendicular components to ensure a straight-line motion along $$AB$$:

$$v_s \sin \theta = v_r \sin 30^\circ$$

Since the magnitudes of both velocities are equal ($$v_s = v_r = v$$), we substitute and cancel out $$v$$ from both sides of the equation:

$$v \sin \theta = v \sin 30^\circ$$

$$\sin \theta = \sin 30^\circ$$

Taking the inverse sine yields:

$$\theta = 30^\circ$$

Conclusion

Double-checking the vector calculation verifies that when the river velocity and swimmer velocity magnitudes are identical, the velocity vectors form a perfect isosceles triangle. This ensures that the heading angle upstream relative to the path matches the path's angle relative to the river bank.

Therefore, the value of $$\theta$$ is 30.