Ship leaves a port at N50°E at 15 knots, then turns 80° southeast 2 hours later. After three hours at 20 knots, what is the bearing of the ship?

Question

Full question: A a ship leaves a port with a bearing of N50°E at a speed of 15 knots. After 2 hours the ship turns 80° southeast. After three hours at a speed of 20 knots, what is the bearing of the ship from the port?

Answer 1

A triangle is formed by the paths taken by the ship.
1 knot = 1.15078 mph

Starting from the origin (0,0)


The first side of the triangle is
2 hours x 15 x 1.15078 mph = 34.5234 miles
at 40 degrees above the x-axis

From there, the ship turns 80 Southeast
and travels 3 x 20 x 1.15078 = 69.0468 miles.
So the second side is 69.0468 miles forming
an 80 degree angle.

IN summary, we have a triangle with sides
34.5234 and 69.0468
having angles 40 degrees and 80 degrees.

The third side is the measure we are trying to find.

It can be found using the Law of Sines formula, which
states the RATIOS of each side and the OPPOSITE angle
are proportional. SPECIFICALLY, a / sin A = b / sin B = c / sin C

So the missing side is X / sin 80 = 69.0468 / sin 40

X = 69.0468 / sin 40 * sin 80 = 105.78