math
posted by sandra on .
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?

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 = 18020 = 160° = N 70° W 
oops
tan x = 6.04/18.23