Motorboats course A motorboat traveled along a triangular co

Motorboat\'s course A motorboat traveled along a triangular course having sides of lengths 2 kilometers, 4 kilometers and 3 kilometers, respectively. The first side was traversed in the direction N20 degree W and the second in a direction S theta degree W where theta degree is the degree measure of an acute angle. Approximate, to the nearest minute, the direction in which the third side was traversed.

Solution

Let\'s call the starting point A, and say the 2 km that went N 20 W is AB; the 4 km in a direction west of south is BC, and CA is the final 3 km. The angle I want to find is angle A.

a^2 = b^2 + c^2 - 2bc cos(A)
Omitting the kilometers (since all four terms are in km^2), I have
16 = 9 + 4 - 12 cos(A)
12 cos(A) = -3
cos(A) = -0.25
Angle A = 104.48 degrees

We know that AB points 20 west of north, and angle CAB is 104.48 degrees. Another way of thinking about AB is, it points 70 degrees north of west. Hence, AC would point 34.48 degrees south of west, or 55.52 degrees west of south; but we were looking for the direction of CA, which is the opposite: 55.52 degrees east of north. That\'s N 55.52 deg E, or N 55 deg 31 minutes E.