A hunter walked into a bush on a bearing of 20° to reach a hut H 600m from his home. He then walked 500m on a bearing of 115° to a trap T, find:
A.) The distance of the trap from his home.
B.) The bearing of the trap from his home.
C.) The bearing of his home from the trap.
D.) The bearing of the hut from the trap.

Given a hut H 600 m at 20° from home, and a trap T 500 m at 115° from the hut, you want distance and bearing of the trap from home, and the bearings of hut and home from the trap.

Vectors

A vector calculator solves this problem nicely, telling us ...

600∠20° + 500∠115° = 746.8∠61.8°

(A, B) The trap is 746.8 m on a bearing of 61.8° from home.

The bearings of home and hut from the trap are 180° more than the bearing of the trap from hut or home.

(C) Home is on a bearing of 61.8 +180 = 241.8° from the trap.

(D) The hut is on a bearing of 115 +180 = 295° from the trap.

Coordinates

Lacking a vector calculator, you can find the coordinates of hut and trap and compute distance and bearing from those.

In (north, east) coordinates, the hut lies at ...

H = 600(cos(20°), sin(20°)) ≈ (563.816, 205.212)

The difference from Hut to Trap is ...

T -H = 500(cos(115°), sin(115°)) ≈ (-211.309, 453.154)

The location of the trap T relative to home is ...

T = H + (T -H) = (563.816, 205.212) +(-211.309, 453.154)

T = (352.507, 658.366)

Then the distance from home is ...

d = √(352.507² + 658.366²) ≈ 746.8 . . . meters

And the bearing from home is ...

θ = arctan(658.366/352.507) ≈ 61.8°

(A, B) The trap is 746.8 m on a bearing of 61.8° from home.

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Additional comment

We like (north, east) coordinates for navigation problems so we can use the usual formulas for vector components and angles unchanged. The vector calculator gives results we can use directly.