A hiker travels N 33 degrees W for 15km, then turns 90 degrees and walks 12km in the direction S 57 degrees W At that time how far is the hiker from his starting point? What is his bearing from his starting point?

Answer 1

As usual, draw a diagram.

the distance is easy, since we have a right triangle

d^2 = 15^2 + 12^2 = 369
d = 19.2

The bearing is N (33+x) W where

tan x = 12/15
x = 37°

bearing is N 70° W

Alternatively, and with much more work,

(0,0) + (15 cos 123°,15 sin 123°) = (-8.17,12.58)
(-8.17,12.58) + (12 cos 213°, 12 sin213°) = (-8.17,12.58) + (-10.06,-6.54) = (-18.23,6.04)

d^2 = 369
tan x = -18.23/6.04 = -3.02
x = 180-20 = 160° = N 70° W

Answer 2

oops
tan x = 6.04/-18.23