Question

A man swims from a point $A$ on one bank of a river of width $100\text{m}$ . When he swims perpendicular to the water current, he reaches the other bank $50\text{m}$ away from point $\text{B}$ downstream. The angle to the bank at which he should swim, to reach the directly opposite point $B$ on the other bank is (in both the cases speed of man with respect to river is same)

• A. ${80}^{\circ }$ with the direction of flow of river
• B. ${110}^{\circ }$ with the direction of flow of river
• C. ${120}^{\circ }$ with the direction of flow of river
• D. ${150}^{\circ }$ with the direction of flow of river

Answer 1

The correct option is C ${120}^{\circ }$ with the direction of flow of river

Let $v_{mw}$ be the velocity of man w.r.t river and $v_w$ be the velocity of river.

From figure $\left(a\right)$, we get
$\tan \theta =\frac{v_w}{v_{mw}}=\frac{50}{100}=\frac{1}{2}$

or $v_{mw}=2v_w$

Using Figure $\left(b\right)$, we can say
$\sin \alpha =\frac{v_w}{v_{mw}}=\frac{v_w}{2v_w}=\frac{1}{2}$

or $\alpha ={30}^{\circ }$
Angle with the direction of flow, $\alpha +{90}^{\circ }={120}^{\circ }$

So, it is ${120}^{\circ }$ with the direction of flow of river.