Question

a ship travels 20miles on a course S 40 degree 10 mins W and then 25miles on a course of N 28 degree 20 mins W . Find the distance and direction of the last position from the frist

Answer 1

make a sketch
I ended up with a triangle with sides 20 and 25 with an angle of 68° 30' between them , or 68.5°
(if you draw in faint vertical and horizontal lines at the critical points, it is easy to calculate angles using the rules of parallel lines. )

So for distance between, use the cosine law to find the missing side
x^2 = 20^2 + 25^2 - 2(20)(25) cos 68.5
..
x = 25.6612..

now find another angle:
sinØ/25 = sin68.5/25.6612..
sinØ = .9064428...
Ø = 65.0° or 115.0°
but since 25 and 25.66 are almost equal , the triangle is almost isosceles, so by common sense I will pick
Ø = 65°

so using my faint vertical lines at the starting and turning points, I can calculate the angle to be N 86°40' W

(to the left of my NS line, I have 3 angles
86°40'
65° and
28°20' , adding up to 180°

distance from starting point = 25.66 miles
direction from starting point = N 86° 40' W