A ship leaves port P and travels 120 miles North to a point X .

Question

A ship leaves port P and travels 120 miles North to a point X. It then turns and travels 60 miles on a bearing of 25 degrees to a point Y.

Draw a diagram of this situation. What angle does the ship’s path make at the point X?  ∘

How far has the ship travelled in total?  miles.

Find the bearing of the ship’s current position (Y) from the port (P).  ∘

Answer 1

going North = bearing of zero degrees

for 120 miles

then a bearing of 25 degrees = an angle of 180-25 = 155 degrees

for 60 miles

connect point Y to the port point, and calculate that line's distance

you have a triangle with SAS, Side Angle Side = 120-155-60

then solve for the side opposite 155 degrees

= 176.2 miles

use laws of cosines and then law of sines

c^2 = 120^2 + 60^2 -2(120)(60)Cos155

c= 176.2

sinP/60 = sin155/176.2

sinP = 60sin155/176.2

P = sin^-1(60sin155/176.2)

= 8.274 degrees bearing

You know c= slightly less than 120+60=180

and the bearing will be slightly more than 0 degrees

3rd side of a triangle has to be less than sum of the other 2 sides

if one angle = 155, the other 2 angles sum to 25 degrees, so bearing is somewhere between 0 and 25 degrees

whatever calculations you come up with have to fit c<180 and 0<P<25