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Sine and Cosine Rule - Bearings

Can I get some help on this question.

A ferry and a cargo ship are both approaching the same port. The ferry is 3.2 km from the port on a bearing of 076° and the cargo ship is 6.9 km from the port on a bearing of 323°.

Find the distance between the two vessels and the bearing of the cargo ship from the ferry.


Here is a diagram which I draw from extracting the info from the question to help me answer this

Is it any correct because it's not helping me, I don't know where to start.

The angle between the horizontal and the line connecting the port and the ferry should be 24, for a start. Can you see why?
Reply 2
I don't understand, how is that.
pretty sure you do this:

angle at cargo is 37 degrees, use the sine rule to get angle at ferry and then use the fact that angle in a triangle add up to 180 and once you got that, use the cosine rule to to get x
Reply 4
generalebriety
The angle between the horizontal and the line connecting the port and the ferry should be 24, for a start. Can you see why?

Really?

uer23
I don't understand, how is that.

The angle between the line of 0 degrees and the horizontal line is 90 degrees, you know the bearing is 76 degrees, so 90-76 = 14.
Reply 5
This should be your diagram:
Reply 6
assmaster
Really?


The angle between the line of 0 degrees and the horizontal line is 90 degrees, you know the bearing is 76 degrees, so 90-76 = 14.


Oh so that means you need a bearing from North to go 76 degrees.
Reply 7
boromir9111
angle at cargo is 37 degrees


I don't understand how this was deduced ( I can't spot and interior/exterior, alternate or any other type of angles).
Reply 8
uer23
I don't understand how this was deduced ( I can't spot and interior/exterior, alternate or any other type of angles).


360-323=37
Reply 9
Thanks a lot for your help guys I have worked it out finally. :smile:
assmaster
Really?

zdo0o
is it because 76 + 24 = 90? :biggrin:

What are you, my primary school? I don't know my times tables either! And I'm proud of it! :mad:

(Yes, 14. Thank you. :o:)
Reply 11
Just need to make sure this is correct as I don't have the answer to this question, can someone check to see if it is right.



1) Using cosine rule to find Side xx

x2=(6.9)2+(3.2)22(6.9)(3.2)cos (113°)x^2 = (6.9)^2 + (3.2)^2 - 2(6.9)(3.2) \cos \ (113°)

x2=75.10468...x^2 = 75.10468...

x=75.10468...x = \sqrt 75.10468...

x=8.67km(3s.f.)x = 8.67km (3 s.f.)

2) Using sin rule to find angle QQ

sin Q6.9\frac{\sin \ Q}{6.9} = sin 113°8.66629...\frac{\sin \ 113°}{8.66629...}

sin Q\sin \ Q = 6.9×sin 113°8.66629...\frac{6.9 \times \sin \ 113°}{8.66629...}

sin Q\sin \ Q = 0.73289...0.73289...

QQ = sin1(0.73289...)\sin^-1(0.73289...)

QQ = 47.12962...°47.12962...°

3) Bearing of cargo ship from ferry

Bearing of cargo ship from ferry = P°+Q°P° + Q°

P°=360°104°P° = 360° - 104°

P°=256°P° = 256°

P°+Q°=256°+47.12962...°P° + Q° = 256° + 47.12962...°

=303.12962...°= 303.12962...°

=303°(3s.f.)= 303° (3 s.f.)
Oh you drew the picture wrong, that makes sense now but yeah your method seems right.
Reply 13
boromir9111
Oh you drew the picture wrong, that makes sense now but yeah your method seems right.


Oh really ? Would you please show me how you would draw the picture.
dude look at your first drawing and compare it to the second, quite a big change there and changes how you solve for it (makes it easier now) but that's what i meant, your first drawing was wrong, pay attention and not lash out on me.
Reply 15
boromir9111
dude look at your first drawing and compare it to the second, quite a big change there and changes how you solve for it (makes it easier now) but that's what i meant, your first drawing was wrong, pay attention and not lash out on me.


:confused: huh? I was not lashing out on you. I thought your comment was referring to my post with the solutions, where you said 'you drew the picture wrong'. So I thought maybe you can show me how it is supposed to be drawn. Be a bit more clear next time.
A cyclist leaves village P and cycles for 25km on a bearing ofv240° to village Q.He leaves Q and cycles for 32km on a bearing of 140° to village R . calculatea) the size of <PQR

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