Question

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?

Answer 1

I hope you know vectors.
let the unit vector along East be  $\hat{i}$.  and the unit vector along north be $\hat{j}$.
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 j  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 i  + 29.69 j    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.