# Question #080f7

Feb 20, 2017 From starting position $O$ the bearing of current is ${345}^{\circ}$ and that of direction of destination is ${100}^{\circ}$. So the angle between the direction of current and the direction of destination is ${\left(360 - 345 + 100\right)}^{\circ} = {115}^{\circ}$

Again the ship is to cover 700km in 20hr to reach at the destination

So its velocity along the direction of destination $O D$ is

${V}_{d} = \frac{700}{20} = 35 \text{km/h}$

The velocity of current along OC

is ${V}_{c} = 5 \text{km/h}$

Le the velocity of ship along OE will be ${V}_{s}$ and this velocity makes an angle $\alpha$ with OD.

Considering cosine law for triangle ODE we can write

${V}_{s}^{2} = {V}_{c}^{2} + {V}_{d}^{2} - 2 {V}_{c} {V}_{\mathrm{dc}} o s 115$

$\implies {V}_{s}^{2} = {5}^{2} + {35}^{2} - 2 \cdot 5 \cdot 35 \cos 115$

${V}_{s} = 37.4 \text{km/h}$

Now applying sin law for triangle ODE we get

${V}_{d} / \sin 115 = \frac{5}{\sin} \alpha$

$\implies \frac{37.4}{\sin} 115 = \frac{5}{\sin} \alpha$

$\implies \sin \alpha = \frac{5}{37.4} \times \sin 115 \approx 0.12$

$\implies \alpha \approx {7}^{\circ}$

So the bearing of the direction of the ship to be driven is ${107}^{\circ}$