# Need help drawing this bearing diagram

• Mar 30th 2008, 07:24 AM
struck
Need help drawing this bearing diagram
A boat sails 10km from harbor H on a bearing of S30E. It then sails 15km on a bearing of N20E. How far is the boat from H? What is its bearing from H?

I think, I only need help with drawing the right diagram for this. I've drawn two, but I guess both were wrong. (Headbang)
• Mar 30th 2008, 08:25 AM
galactus
One way is to use the law of cosines. We have an angle and two sides.

The angle is 50 degrees and the two sides are 10 and 15 km.

The distance from the ship to the harbor can be gotten from:

$a=\sqrt{10^{2}+15^{2}-2(10)(15)cos(50)}\approx{11.50} \;\ km$

Once we have that distance, the bearing back to the harbor can be gotten from the law of sines:

$\frac{11.5}{sin(50)}=\frac{10}{sin(B)}\Rightarrow{ \frac{10sin(50)}{11.5}}$

$B=41.785 \;\ degrees$

Once we get that angle add 200 degrees to get your azimuth.

We find it is about 242 degrees. That is a bearing of

$S62W$
• Mar 30th 2008, 09:39 AM
struck
Thanks. But currently, I can neither use cosine rules, nor do I know them. Is there any way to solve it using basic trigonometry and angles/triangles laws?
• Mar 30th 2008, 10:23 AM
galactus
I don't know what you have to use, but there are many ways to go about it.

Let the coordinates of the harbor be (100,100).

Then the coordinates of the first point the ship sails to is

$x=100+10sin(150)=105$

$y=100+10cos(150)=91.34$

Let's go to the second point from there.

Its coordinates are:

$x=105+15sin(20)=110.13$

$y=91.34+15cos(20)=105.435$

The distance back to the harbor is then the distance formula:

$D=\sqrt{(110.13-100)^{2}+(105.435-100)^{2}}=11.50$

The bearing back to the harbor is:

$tan^{-1}(\frac{110.13-100}{105.435-100})=61.785 \;\ deg$

Which is a bearing of about $S62W$

As before.