Question

19-21. A boat sail from A on a bearing of 140 for 40 km to B. Then sails on a bearing of 250 for 80 km to C.
Find distance AC.
Find the bearing of A from C.
Find the bearing of C from A.

Answer 1

make your sketch to see a triangle ABC with angle B = 70°
AC^2 = 40^2 + 80^2 - 2(40)(80)cos70°
= 1600 + 6400 - 2188.9289..
AC = √ .....
= appr 100.94

Use your diagram and this result to find the angles needed

Answer 2

draw a diagram, and then use the law of cosines to find AC
In ∆ABC, B = 70°, so
b^2 = 40^2 + 80^2 - 2(40)(80) cos70°
if A is at (0,0) then
B is at (25.71,-30.64)
C is at (-49.46,-58.00)
so the bearing of A from C is θ, where
tanθ = 58/49.46
making the bearing of C from A θ+180°