A surveyor leave her base camp and drive 42 km on a bearing of 032, she then drive 28km on a bearing of 154.How far is she then from her base camp and what's her bearing from it

Step-by-step explanation:

The scenario is represented in the attached photo. Triangle ABC is formed. AB represents her distance from her base camp. We would determine BC by applying the law of Cosines which is expressed as

a² = b² + c² - 2abCosA

Where a,b and c are the length of each side of the triangle and B is the angle corresponding to b. It becomes

AB² = AC² + BC² - 2(AC × BC)CosC

AB² = 42² + 28² - 2(42 × 28)Cos58

AB² = 1764 + 784 - 2(1176Cos58)

AB² = 2548 - 1246.37 = 1301.63

AB = √1301.63

AB = 36.08 km

To find the bearing, we would determine angle B by applying sine rule

AB/SinC = AC/SinB

36.08/Sin58 = 42/SinB

Cross multiplying, it becomes

36.08SinB = 42Sin58

SinB = 42Sin58/36.08 = 0.987

B = Sin^-1(0.987)

B = 81°

Therefore, her bearing from the base camp is

360 - 81 = 279°

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