# Bearings Problem

I'm presented with the following bearings problem. I believe I have graphed it correctly, although I don't know where to go from here.

A US Coast Guard patrol boat leaves Port Cleaveland and averaged 35 knots (nautical mph) traveling for 2 hours on a course of 53 degrees and then 3 hours on a course of 143 degrees.

I need to find the boat's bearing and distance from Port Cleaveland.

• Won't you calculate the degrees from the east direction?. If you did, it is simple application of pythogorus theorem. It travels 2*35 nm at 53 degress to the east. Then it travels 3*35 nm at 143 degrees to the east. It's position is $\sqrt{70^2 + 105^2} = 126.2$ miles from the port of cleaveland and it is 109.3 degrees to the east. Commented Mar 24, 2014 at 1:46

Won't you calculate the degrees from the east direction?. If you did, it is simple application of pythogorus theorem. It travels 2*35 nm at 53 degress to the east. Then it travels 3*35 nm at 143 degrees to the east. It's position is $\sqrt{70^2 + 105^2} = 126.2$ miles from the port of cleaveland and it is 109.3 degrees to the east.

The distance would be the length of the hypotenuse in the right triangle. You can find the legs by multiplying speed*time. Use the Pythagorean Formula $a^2+b^2=c^2$.

Just as a side note, I do not think this problem has an answer. It only says that the average speed was 35 knots, so, for example, the boat could have spent the first two hours hurtling down the 53 degrees course, and then just sat there for the remaining three hours.

Let A = location of Port Cleveland, B = location of boat after 2 hours on course 53 degrees,

and C = final destination. Then triangle ABC is a right triangle at B since B = 360 - 127 -

143 = 90 degrees. Further AB = 2*35 = 70 mi, BC = 3*35 = 105 mi. So apply Pythagorean theorem

for this triangle: AC = (AB^2 + BC^2)^1/2 = (70^2 + 105^2)^1/2 = 126.2 mi. We next find angle

BAC. sin(BAC) = 105/126.2 = 0.832. So BAC = sin^(-1)(0.832) = 56.3 degrees. Thus the bearing

of the boat is: 53 + 56.3 = 109.3 degrees, and its distance from Port Cleveland is: 126.3 mi.