Boat A is 20 km from port on a bearing of 025° and boat B is 25 km from port on a bearing of 070°. Boat B is in distress. What bearing should boat A travel on to reach boat B?

Respuesta :

Answer: Boat A should travel 152.8° to reach B

Step-by-step explanation:

The diagram illustrating the scenario is shown in the attached photo. Triangle ABP is formed. A represents the position of boat A. B represents the position of boat B. P represents the position of the port.

We would determine AB by applying the law of cosines

AB² = AP² + BP² - 2AP×BPCosP

AB² = 20² + 25² - 2 × 20 × 25 × Cos45

AB² = 1025 - 707.10678 = 317.89322

AB = √317.89322 = 17.83

We would determine the bearing of B from A by finding angle A. We would apply the sine rule.

AB/SinP = AP/Sin A

17.83/Sin45 = 20/SinA

Cross multiplying, it becomes

17.83 × SinA = 20Sin45 = 14.14

SinA = 14.14/17.83 = 0.79

A = Sin^-1(0.79) = 52.2°

The total angle at A is 65 + 52.2 = 117.2°

The angle formed outside the third quadrant is 117.2 - 90 = 27.2°

Therefore, bearing B from A is

180 - 27/2 = 152.8°

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