From a resting place, a bushwalker hikes due north for 1.5km to a waterhole and then on a bearing of 315° for 2km to base camp. a find how far west the base camp is from the waterhole, to the nearest metre. b find how far north the base camp is from the waterhole, to the nearest metre. c find how far north the base camp is from the initial resting place, to the nearest metre.

The attached figure shows formulation of the problem in this question.

(a) Base camp to the waterwhole to the West
From the figure, this is distance DB.
Angle CBD = 45°
Therefore,
Cos 45 = DB/2 => DB = 2 Cos 45 = 1.41421 km = 1414.21 m
To nearest metre,
DB = 1414 m

(b) Base camp to the waterhole to the North
From the figure, this is CD.
Therefore,
Sin 45 = CD/2 => CD = 2 Sin 45 = 1.41421 km = 1414.21 m
To nearest meter,
CD = 1414 m

(c) Base camp to the initial resting place to the North
From the figure, this is CE.
From geometry;
CE = CD+ DE, but DE = AB = 1.5 km = 1500 m
Then,
CE = 1414+1500 = 2914 m

Dis the solution here