A ship sails N 27°10' E for 39 km and then turns N 67°50' for 85 km. What isad the distance and the bearing of the ship from the starting point?

Question

>>>So, I have a sketch of one triangle(not a right triangle, then drew 3 right triangles all over it then solved for their sides and angles..

I came with a distance of 90.38 km, a bearing which is N 47.63° W. Did I got them right?

Answer 1

I assumed that by N 67°50' , you meant
N 67°50' W
I had a triangle with sides 85 and 39 and the angle between them as exactly 85°
Using the cosine law I got your distance.
Good job

now look at the actual triangle.
let the angle at the bottom be Ø
so sinØ/85 = sin85/90.38
sinØ = .93689...
Ø = 69.536 , but that includes our original 27° 10'

so our bearing will be 69.536° - 27°10'
= 69°32' - 27°10'
= 42° 22'

or N 42°22' W

Here is a little hint:
using your D°M'S key on your calculator you can actually perform all arithmetic calculations in the degree-minute-second notation
Play around with it a bit to become familiar with it.
So unfortunate that degree measurements and time measurements are not metric.