A swimmer can swim with velocity of 12 km/h in still water. Water flowing in a river has velocity 6 km/h. The direction with respect to the direction of flow of river water he should swim in order to reach the point on the other bank just opposite to his starting point is ________°. (Round off to the Nearest Integer) (find the angle in degree)

We are given a swimmer with speed $$v_s = 12$$ km/h in still water and a river flowing with speed $$v_r = 6$$ km/h. The swimmer wants to reach the point directly opposite on the other bank, so the net displacement must be purely perpendicular to the river flow.

Let the swimmer swim at an angle $$\theta$$ measured from the upstream direction of the river flow. The swimmer's velocity has two components: along the river, $$v_s \cos\theta$$ (upstream), and perpendicular to the river, $$v_s \sin\theta$$ (across). For the swimmer to reach the point directly opposite, the net velocity along the river must be zero. The river carries the swimmer downstream at $$v_r = 6$$ km/h, and the upstream component of the swimmer's velocity is $$v_s \cos\theta$$. Setting these equal:

$$v_s \cos\theta = v_r$$

$$12 \cos\theta = 6$$

$$\cos\theta = \frac{1}{2}$$

$$\theta = 60°$$

This angle $$\theta = 60°$$ is measured from the upstream direction. But the question asks for the angle with respect to the direction of flow of the river, which is the downstream direction. The angle from the downstream direction to the swimmer's direction (going upstream and across) is $$180° - 60° = 120°$$.

To verify: the perpendicular component of velocity is $$12 \sin 60° = 12 \times \frac{\sqrt{3}}{2} = 6\sqrt{3}$$ km/h (crossing the river), and the along-river component is $$12 \cos 60° = 6$$ km/h upstream, which exactly cancels the 6 km/h downstream river flow. The net velocity is purely perpendicular, confirming the swimmer reaches directly opposite.

The required angle with respect to the direction of flow is $$\boxed{120}$$°.