a ship leaves point A and travels 60 miles due east, then travels northward 59 degrees to point C, and travel 60 miles to point C, find the distance from point A to point C

Answer 1

I will interpret "northward 59° " as
N31°E

Since the two distances are the same, we have an isosceles triangle and the angles at A and C would each be 29.5°
let x = (1/2)AC ----< AC = 2x
cos 29.5 = x/60
x = 60cos29.5 = 52.22..
so AC = appr 104.44 miles

or, using the cosine law
d^2 = 60^2 + 60^2 - 2(60)(60)cos 121°
= 10908.27...
d = √10908.27..
= 104.44