help me spot my mistake! maths trigonometry

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Mia 15
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Q12) A ship travelling with a constant speed and direction and direction is sighted from a lighthouse. At this time it is 2.7km away on a bearing of 042. Half an hour later it is on a bearing of 115 at a distance of 7.6km from the same lighthouse. Find its speed in kilometres per hour.

I drew a diagram and used the formula c^2 = a^2 + b^2 - 2abcosC
so c^2 = (2.7)^2 + (7.6)^2 - 2(2.7)(7.6)cos(115)
c^2 = 82.4
c = 9.1

9.1/0.5 =18.2 km / h

which isn't correct because the answer says 14.6 km/h
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Notnek
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(Original post by Mia 15)
Q12) A ship travelling with a constant speed and direction and direction is sighted from a lighthouse. At this time it is 2.7km away on a bearing of 042. Half an hour later it is on a bearing of 115 at a distance of 7.6km from the same lighthouse. Find its speed in kilometres per hour.

I drew a diagram and used the formula c^2 = a^2 + b^2 - 2abcosC
so c^2 = (2.7)^2 + (7.6)^2 - 2(2.7)(7.6)cos(115)
c^2 = 82.4
c = 9.1

9.1/0.5 =18.2 km / h

which isn't correct because the answer says 14.6 km/h
The angle in your working is wrong. Notice that you haven't used the "042" anywhere so that can't be right.

Please post a pic of your diagram.
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Mia 15
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(Original post by Sir Cumference)
The angle in your working is wrong. Notice that you haven't used the "042" anywhere so that can't be right.

Please post a pic of your diagram.
here's the diagram I drew, sorry if its hard to see
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GabiAbi84
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Remember bearings always are measured from the north line so both points bearings should be measure from north.
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Mia 15
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(Original post by GabiAbi84)
Remember bearings always are measured from the north line so both points bearings should be measure from north.
ohh thanks I see what I did wrong!
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