# Trigonometry problem involving oblique triangle

How would I solve the following problem?

A ship sails $15$ miles on a course $S40^\circ10'W$(south 40 degrees 10 minutes west) and then $21$ miles on a course $N28^\circ20'W$(north 28 degrees 20 minutes west). Find the distance and direction of the last position from the first.

I think that to solve this problem I must make an oblique triangle. One side of the triangle would be $15$ and have an angle of $40^\circ10'$ and the other side would be $21$ and have a angle of $28^\circ20'$.

I think what I have to find is the side which connect these two sides.

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I would draw a diagram - I find that generally clarifies such questions. – Mark Bennet Aug 15 '12 at 15:21

Let $P$ be our start point, $Q$ where we change course, and $R$ the end point. First choose $P$. Then draw a North-South thin line through $P$ as a guide. To reach $Q$ we turned $40^\circ 10'$ clockwise from due South, and sailed $15$ km. Draw $Q$. Draw a thin North-South line through $Q$ as a guide. We sail $21$ km in a direction $28^\circ 20'$ counterclockwise from due North. Draw the point $R$ that we reach.
By properties of transversals to parallel lines (our two guide lines), $$\angle Q=40^\circ 10'+28^\circ 20'=68^\circ30'.$$
We can now find the required distance $PR$ we are from our start position by using the Cosine Law. For $$(PR)^2=15^2+21^2-2(15)(21)\cos 68^\circ30'.$$ (I get $PR\approx 20.86$, but my calculations are not to be trusted.) Now for the direction. We will know everything once we know $\angle A$. For this, we could use the Cosine Law, but the Sine Law is easier. We have $$\frac{\sin A}{21}=\frac{\sin 68.5^\circ}{PR}.$$ I get (but again don't trust me) that $A\approx 69.5^\circ$.
If we use the North-South guideline, our position at $R$, as viewed from $P$ is obtained by facing due South and turning clockwise through about $40^\circ10'+69^\circ 30'$. To express this in the notation that your problem was put, subtract from $180^\circ$. We get $70^\circ 20'$. So $R$ is North, $70^\circ 20'$ West from $P$.