stuck with: bearings?

Question

2 ships leave port at exactly the same time.
Ship A travels at 12 kilometres per hour on a bearing of 073 degree.
Ship B sets out at 15 kilometres per hour.

After 3 hours the two ships are 30 km apart.

Calculate, to the nearest degree, the bearing Ship B must have followed when leaving port.

I am really stuck with this one, must be late , or maybe the time change!! ;-) Help really appreciated!!

Katrina

Answer 1

There is not a unique answer. It is either 031 or 115 depending on whether ship B headed further north or south of ship A

Answer 2

It's really much easier to solve this if you make a rough sketch first.
Say the port is the "origin." The first ship travels 36 km at a heading of 73° - that's a point at polar coordinates of 36, angle 17. (because polar coordinates are measured CW from the x-axis, but bearings are measured CCW from the y-axis)

You don't know where the second ship is, but you know it traveled 45 km - draw a circle of radius 45 centered on the origin. The second ship is somewhere on the circumference of that circle.

The ships are 30 km apart - can you guess what to do? Draw a circle of radius 30 centered on the first ship.

The circles intersect at 0, 1 or 2 points, and the intersection points (if any) define the possible locations of the first ship. The intersection points will also let you construct the necessary triangles to compute the location of the second ship.

Answer 3

the bearing of 073 degrees is not of great importance as such but rather the ships' separation angle.

you can form this problem into a triangle as such .....A travels 36km, B travels 45km in 3hrs, therefore draw two lines starting from the same point separated by some angle ( call it D). line 1 will be 36 long, line 2 will be 45 long. you then join the tips of the lines and label this length 30 ( your separation). label the remaining angles E and F.

now the tricky but simple bit...

Use the Cosine Rule for triangles...
(d^2) = (e^2) + (f^2) - 2efCosD

now we want to calculate the starting separation angle and we already know the outer lengths.

on your triangle diagram: the side opposite angle D is 30 long, and depending on your labelling the sides opposite E and F are 45 and 36 respectively. so if a side is opposite angle D its denoted with little d in the equation above.e.g side opposite angle F is denoted f, etc

now just put the numbers in...

(30^2) = (45^2) + (36^2) - 2(45x36)CosD

you get

CosD = 0.7472
therefore D = 42 degrees.....

As long as the ships set off at their respective speeds separated by approx. 42 degrees, then after 3 hrs they will be 30km apart.....

Happy bearings......

Answer 4

Well I agree with saljegi about the triangle sides of 30,36,and 45 and I undestand that it could be either bearing N or S of 73°, but I've long forgotten how to do cosines! If you know, this might help. Far too complicated for me!

http://lasp.colorado.edu/~bagenal/MATH/supplement/lawofcosines.html

Answer 5

Draw it to scale. I think you might have to use the sine rule to solve this, you have a unique triangle with sides 30km,45km and 36km.