Two indications for bearings?!

Hi everyone!

So I kind of have a huge problem... We started with trigonometry two weeks ago and now reached bearings. I've done perfectly fine with it, as long as it only contained *1 bearing *thingy. But in this exercise we had to use two of them.

Here's the task:

A yacht sails 15 km on a bearing of 053° and then 7 km on a bearing of 112°.

How far north is the yacht from its starting position.

See, I understood how I had to calculate the first part. But then after I got that (how far north it goes with a bearing of 053°), I couldn't seem to figure out how I had to construct a sketch and then calculate it to the correct result. Solution would be 6.40km

I would be so greatfull if any of you could help me get to that result!

Thanks in advance,

Joelle ((:

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Re: Two indications for bearings?!

Re: Two indications for bearings?!

Quote:

Originally Posted by

**trigophobic** Hi everyone!

So I kind of have a huge problem... We started with trigonometry two weeks ago and now reached bearings. I've done perfectly fine with it, as long as it only contained *1 bearing *thingy. But in this exercise we had to use two of them.

Then do it in two parts- use the information given for the first leg to find where the yacht is at the end of the first leg, then use the information given for the second leg, starting from the position you just found, to find the final position. Technically, that would be finding the "displacement vector" for each leg, then adding the two vectors.

There is a way to do it without using "vectors" or doing the two legs separately, but it is, admittedly, harder. If you look at the picture skeeter drew, notice that the NS, EW, first leg lines form a right triangle. Since the first angle given is 53 degrees (on the lower left), the other acute angle on that right triangle (on the upper right) is 90- 53= 37 degrees. Also the angle between the south line and the new leg is 180- 112= 68 degrees. If you draw a line from the beginning point to the ending point (which is, after all the information we want to find), we have a triangle with two sides of length of 15 and 7 and angle between those sides of 37+ 68= 105 degrees. Now you can use the "cosine law", "if a triangle has sides of length a, b, c, and the angle opposite side c has measure C, then $\displaystyle c^2= a^2+ b^2- 2ab cos(C)$: to find the length of the third leg. You can use the sine law", "if a triangle has sides of length a, b, and c with opposite angles A, B, and C, respectively, then $\displaystyle \frac{sin(A)}{a}= \frac{sin(B)}{b}= \frac{sin(C)}{c}$.

Quote:

Here's the task:

A yacht sails 15 km on a bearing of 053° and then 7 km on a bearing of 112°.

How far north is the yacht from its starting position.

See, I understood how I had to calculate the first part. But then after I got that (how far north it goes with a bearing of 053°), I couldn't seem to figure out how I had to construct a sketch and then calculate it to the correct result. Solution would be 6.40km

I would be so greatfull if any of you could help me get to that result!

Thanks in advance,

Joelle ((: