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?

Question

James Adkins · Accepted Answer

About #111.3^circ#

The bearing is the angle measured clockwise from true north.

It will be #73^circ# 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:

#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#

Oliver Atkinson · Answer

The problem is inconsistent; adding the vectors we get about #91^circ.#

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^circ# is equal to #90^\circ-73^\circ=17^\circ# 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^circ# is #b# and opposite is #a#. So

# a = 63 \ sin 73^circ#

#b = 63 \ cos 73 ^circ#

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

#c = a + 60 sin 110^circ = 63 \ sin 73^circ + 60 sin 110^circ approx 116.6#

#d = b + 60 \ cos 110 ^circ = 63 \ cos 73 ^circ + 60 \ cos 110 ^circ approx -2.1#

If #c# was zero that would be a bearing of #90^circ;# it's a bit more at

#theta =180^circ + arctan ( 116.6/ -2.1 ) approx 91^circ #

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^circ# consistent with my calculation.

Looks like a bad problem.