Question

A ship sails 30km from A on a bearing of 048 degrees and a further 42km on a bearing of 110 degrees to arrive at B. What is the distance and bearing B from A.

Answer 1

The distance from A to B is approximately 67.5 km

The bearing of B from A is approximately 86.89°

Explanation:

i) The distances and the bearing of the motion of the ship is given as follows;

The distance the ship sails on a bearing of 048° = 30 km

The distance further the ship sails on a bearing of 110° = 42 km

From the drawing of the sailing of the ship created with Microsoft Visio, we have;

∠C = 48° + 70° = 118°

By cosine rule, we have;

c² = b² + a² - 2·b·a·cos(∠C)

From which we have;

c² = 30² + 48² - 2 × 30 × 48 × cos(118°) ≈ 4,556.07810082

c ≈ √(4,556.07810082) ≈ 67.5

The distance from A to B = c ≈ 67.5 km

ii) By sine rule, we have;

a/sin(A) = b/sin(B) = c/sin(C)

∴ sin(A) = a·sin(C)/c

Substituting the values of the variables gives;

sin(A) ≈ 48·sin(118°)/67.5

∠A ≈ arcsin(48·sin(118°)/67.5) ≈ 38.89°

Therefore, the bearing of B from A ≈ 48 + 38.89 = 86.89°.