Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
1  
I would draw a diagram - I find that generally clarifies such questions. –  Mark Bennet Aug 15 '12 at 15:21

1 Answer 1

up vote 3 down vote accepted

The distances sailed are small enough that the curvature of the Earth makes no significant difference and we can use plane trigonometry.

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 $P$.

share|improve this answer
    
Yes your response is excellent and correct. –  Fernando Martinez Aug 15 '12 at 16:51

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.