Question #080f7

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Question #080f7

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Answer 1 · 2017-02-20T09:28:53

drawn

From starting position $O$ the bearing of current is $345^@$ and that of direction of destination is $100^@$ . So the angle between the direction of current and the direction of destination is $(360-345+100)^@=115^@$

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

So its velocity along the direction of destination $OD$ is

$V_d=700/20=35"km/h"$

The velocity of current along OC

is $V_c=5"km/h"$

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

$V_s^2=V_c^2+V_d^2-2V_cV_dcos115$

$=>V_s^2=5^2+35^2-2*5*35cos115$

$V_s=37.4"km/h"$

Now applying sin law for triangle ODE we get

$V_d/sin115=5/sinalpha$

$=>37.4/sin115=5/sinalpha$

$=>sinalpha=5/37.4xxsin115~~0.12$

$=>alpha~~7^@$

So the bearing of the direction of the ship to be driven is $107^@$

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Question #080f7

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