About 111.3∘
Explanation:
The bearing is the angle measured clockwise from true north. It will be 73∘ more than the triangle's angle opposite 60 km. The problem seems overdetermined, meaning there are several different ways to get the answer that might not give the same result. We'll use the Law of Cosines: 602=632+952−2(63)(95)cosθcosθ=632+952−6022(63)(95)=671855θ≈38.3∘ Bearing of 73∘+38.3∘=111.3∘
Dean R.
Jun 28, 2018
The problem is inconsistent; adding the vectors we get about 91∘.
Explanation: Here's a solution that doesn't use the Law of Cosines.
Let's put this on the Cartesian Grid with the port being the origin
(0,0)and North indicated by the positive y axis, east the positive x axis in the usual way. Bearings are complementary to the usual angles we consider with the x axis, so a bearing of 73∘ is equal to 90∘−73∘=17∘ relative to the positive x (eastward) axis.
But we don't have to think about it that way; sine and cosine kinda reverse when we do things relative to the y axis, but it's not that complicated.
Let's call the spot after the first leg, 63 km, point A(a,b).
The points (0,0),(0,b) and(a,b) make a right triangle whose adjacent to 73∘ is b and opposite is a. Soa=63 sin73∘b=63 cos73∘ It's switched from normal because we're using bearings. c=a60sin110∘=63sin73∘+60sin110∘≈116.6d=b+60cos110∘=63cos73∘+60 cos110∘≈ −2.1 If c was zero that would be a bearing of 90∘; it's a bit more atθ=180∘+arctan(116.6−2.1)≈91∘ That's inconsistent with the answer determine by the Law of Cosines, either because I made a mistake or the person who wrote the question made a mistake. It's also inconsistent with OA=95, it's around 117 this way. Let's see if we can figure out which. Alpha knows. Note it says edge lengths 63, 60, 116.648. Compare that to Those are not the same triangles but it looks like my math is OK. Angle A is 180−(90−73)+(110−90)=143∘ consistent with my calculation. Looks like a bad problem.:(
