find the bearings for the last two legs of the race. (Round your answer to one decimal place.)

Question

A boat race runs along a triangular course marked by buoys A, B, and C. The race starts with the boats headed west for 3600 meters. The other two sides of the course lie to the north of the first side, and their lengths are 1700 meters and 3200 meters. Draw a diagram that gives a visual representation of the problem. Then find the bearings for the last two legs of the race. (Round your answer to one decimal place.)

N	° E
S	° E

Answer 1

Use the Law of Cosines

36^2 = 17^2 + 32^2 -2(17(32)Cos(C) where C=the angle opposite AB=c=36

Cos(C) = (36^2-17^2-32^2)/-2(17)(32)

then use the law of Sines to find one other angle, either A or B

then add those two angles and subtract from 180 to get the 3rd angle

then convert the angles to bearings

after some tedious calculation, with a calculator, it looks like about 25, 66 and 89.1 degrees which don't sum to exactly 180, due to rounding approximations.

or use one of the online triangle calculators where you just punch in the 3 sides and out pops the 3 angles

62.7, 28.1 and 89.2 seems to be the solution on one website, suggesting an error in the above tedious calculations, or try another triangle calculator on another website and see if it agrees

the conversion to bearings will be off a little from what's below based on 25, 66 and 89.1

due west for the first leg of the triangle

then North 0.9 degrees East for 3200 miles

then South 24 degrees East for 1700 miles

or if the legs are switched

due West for 1st leg

then North 24 degrees East for 1700 miles

then South 0.9 degrees East for 3200 miles