Question

A river is flowing with a speed of $1\text{km/hr}$ as shown in figure. A swimmer wants to go to the point C starting from point A. He swims with a speed of $5\text{km/hr}$ with respect to the flowing river at an angle $\theta$ as shown. If $AB=BC=400\text{m}$, then value of $\theta$ is:

• A. ${37}^{\circ }$
• B. ${30}^{\circ }$
• C. ${53}^{\circ }$
• D. ${45}^{\circ }$

Answer 1

The correct option is A ${37}^{\circ }$

Given : $AB=BC=400\text{m}=0.4\text{km}$
$v_m=v_{mr}+v_r$..................(1)
where $v_m=\text{velocity of swimmer w.r.t gound}$
$v_{mr}=\text{velocity of swimmer w.r.t river}=5\text{km/hr}$


${v_m}_x=(v_{mr}\cos \theta +v_r)$
$=(5\cos \theta +1)\text{km/hr}$......(2)
Similarly:
${v_m}_y=(v_{mr}\sin \theta +0)$
$=5\sin \theta \text{km/hr}$...........(3)

$\therefore$ Time taken by swimmer to cross the river is given by:
$t=\frac{AB}{v_{my}}=\frac{0.4}{5\sin \theta }$
$\Rightarrow$ Distance covered by swimmer in time interval $t$
$BC={v_m}_x\times t=(v_{mr}\cos \theta +v_r)t$
i.e $0.4=(5\cos \theta +1)\times \frac{0.4}{5\sin \theta }$
$\Rightarrow 5\sin \theta -5\cos \theta =1$
On squaring both sides:
$25{\sin }^2\theta +25{\cos }^2\theta -50\sin \theta \cos \theta =1$
$\Rightarrow 25-25\sin 2\theta =1$
$\Rightarrow 25\sin 2\theta =24$
$\Rightarrow 2\theta ={\sin }^{-1}\frac{24}{25}={73.74}^{\circ }$
$\therefore \theta \simeq {37}^{\circ }$