# Problem 1 of 3

A ship travels 15 km on a bearing of 070°. It then changes direction and travels 22 km on a bearing of 200°.

(a) Draw a vector diagram modelling this situation.

(b) How far is the ship from its starting point?

(c) Determine the direction the ship must head in order to return to its starting point. Answer to the nearest degree.

**Solution**

(a)

(b) To solve this problem we can use the sin law or the cosine law to find the distance of AC(A is the starting point, B is the point where they changes direction and C is the final position of the ship).

**Sine Law Solution**

a/Sin A = b/Sin B

22 Km/Sin 87.08° = b/Sin 50°

b*Sin 87.08° = 22 km*Sin50°

b = (22 km*Sin50°)/Sin87.08°

b = 16.87 km

**Cos Law Solution**

b^{2} = a^{2}+c^{2}-2*a*c*Cos B

b^{2} = (22^{2}+15^{2})-(2*22*15*Cos 50°)

b^{2} = 709-424.23

b^{2} = 284.77

b = 16.87 km

**The ship is 16.87 km away from the starting point**

(c) There are 3 ways to anwers this question, we can use the bearing method, the degrees and compass method or the compass and degrees method.

**Using the Bearing Method**

The ship must head 16.87 km at a bearing of 292.92° or at 293°

**Using the Degrees and Compass Direction Method**

22.92° north of west or 67.08° west of north

**Using the Compass Direction and Degrees**

North 22.92° West or West 67.08° North

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