# A ship leaves a port on a bearing of 073° and sails 63km. The ship then changes course and sails a further 60km on a bearing of 110° where it anchors. When it anchors it is 95km from the port. Calculate the bearing of the ship from the port at this point?

About

It will be #73^circ# more than the triangle's angle opposite 60 km.

#60^2 = 63^2 + 95^2 - 2 (63)(95) cos theta#

#cos theta = {63^2 + 95^2- 60^2}/{2(63)(95)} = 671/855#

#theta approx 38.3^circ#

Bearing of #73^circ + 38.3^circ = 111.3^circ#

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The problem is inconsistent; adding the vectors we get about

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

Bearings are complementary to the usual angles we consider with the x axis, so a bearing of

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

The points

It's switched from normal because we're using bearings.

If

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

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

Looks like a bad problem.

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