# A shop leaves its port and travels on a bearing of 075 for 11km. It then turns and travels due north for 32km, after which it changes course to a bearing of 259, travels for a further 27km and then stops. i) How far is the ship from its port? ii) What is the bearing of the port from the ship?

1

2015-04-07T19:38:17+05:30

### This Is a Certified Answer

Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
I hope you know vectors.
let the unit vector along East be  .  and the unit vector along north be .
The bearing is the direction specified from the North direction along the clockwise direction.

First leg of the travel: bearing of 075 for 11 km:
displacement vector is :  11 [ Cos(90-75) i + Sin (90-75) j ] km
second leg of the travel: bearing of 0  for 32 km:
displacement vector is:  32 km
third leg of the travel is:  259 for 27 km.
angle between East & direction of travel: 259-90= 169 deg in clockwise direction.
displacement vector is:   27 (Cos (-169) i + Sin (-169) j ) km

Final displacement : sum of the three vectors;
=  [11 Cos 15 + 27 Cos 169] i + [ 11 Sin 15 + 32 - 27 Sin 169 ] j
=  -15.88 + 29.69   km
√(15.88² + 29.69²) = 33.67
Cos Ф = -15.88 / 33.67 = -0.4716
Sin Ф = 29.69 / 33.67
Ф = 118.14°  from East anti clockwise. So 18 deg from North.

final position:
So bearing of the ship from the port  is 360 -  18 = 342 from north.
So bearing of the port from the ship is : 180 - 18= 162 from north.

Distance from the port: 33.67 km.

sir you should know trignometric values for this
they will give in question but we should simplify to multiples and sub multiples
please have a look at my doubts