Please help with this question
I tried working it out, but I couldn't get it at all

2 ships leave port. The first ship sails on a bearing of 270 degrees for 8 km while the second ship sails on a bearing of 200 degrees for 8 km. by drawing, find
(a) the distance between the ship
(b) the bearing of the first ship from the second
(c) the bearing of the second ship from the first

Maybe try drawing out the problem?

I did I couldn't get it

Ahh is there a formula for it? I am trying to think back to Math class but I think we had a formula to figure out this problem.

c2= a2+b2 - 2(a)(c)Cos x?

or side a / sin a?
side b/ sin b?
side c/ sin c?

This website might help you
Calculate distance and bearing between two Latitude/Longitude points using Haversine formula in JavaScript

Do you have the answers for a and b?

No I don't :(

It doesn't help
I need working!

Well, this can be solved mathematically, but the question as presented asks for a drawing. It is easiest to do with circular graph paper, but you can simulate that on regular paper if you are very careful in your measurements and use a large enough scale so that the thickness of your lead (might sound funny, but I don't mean it as a joke) doesn't degrade the accuracy of your answer.

If you use a scale something like 2 cm = 1 km you should be OK. Both ships start at the port, which would be the center of the circular graph. Draw one line 16 cm straight west (270 degrees). (or pick a different scale). Draw another line the same length at 200 degrees - which is SSW. The ends of the lines will give you the relative positions of the ships.
Measure the distance between the two positions. Apply your scale and you have the distance between the ships. For the bearing of ship 1 from ship 2, measure the angle offset from North.
With circular graph paper, you can use a parallel ruler lined up on the positions, slide to the center and read the bearing at the upper left part of the circle. On regular paper, there are a number of ways you can do it, but I think the easiest is to drop a perpendicular line from the course of ship 1 such that the line passes through ship 2's position. At the position of ship 2, measure the angle between this line and a line between the ship positions.

Subtract this from 360 degrees to get the bearing. Expect an answer somewhere around 350 degrees.

For the bearing of ship 2 from ship 1, either plot the logical opposite, or simply subtract 180 degrees

And the distance?
I'm not getting that either

I don't know which angle to use in the cosine rule :(

Which angle should I use in the cosine rule galactus?

The law of cosines is used when you have SAS.

You can see from the drawing, the angle needed is 70 degrees?

So, the distance between the ships is:



Also, you can use the law of sines to find the other angles to aid in the other two parts of the problem.

Once you have the above solution, to find the angle at ship A use:



then solve this for A to find that angle.

In other words, solved for angle A, it is

One thing to note is that this is an isosceles triangle, therefore, angle B will be the same as angle A

Once you have those, parts b and c are easy.

Are you sure galactus?

How did u get 70 degrees in the first place man?

How did I get 70 degrees? Come on. What is 270-200?

Oh Im sorry
I'm not smart like you... so yeah haha


What about the
The bearing of the first ship from the second and
The bearing of the second ship from the first

How do I do that?

Galactus please help me out!

What about the
The bearing of the first ship from the second and
The bearing of the second ship from the first

How do I do that?

Look more carefully at Galactus' post

Quote:

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Originally Posted by galactus

The law of cosines is used when you have SAS.

You can see from the drawing, the angle needed is 70 degrees?.

So, the distance between the ships is:



Also, you can use the law of sines to find the other angles to aid in the other two parts of the problem.

Once you have the above solution, to find the angle at ship A use:



then solve this for A to find that angle.

In other words, solved for angle A, it is

One thing to note is that this is an isosceles triangle, therefore, angle B will be the same as angle A

Once you have those, parts b and c are easy.

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