Qualification > Reference Material
Add Maths June 8, 10 2010
holtadit:
--- Quote from: J.Darren on May 30, 2010, 01:37:56 pm ---The ship is travelling at a bearing of 090, and it is 90km away from the lifeboat at the time of the call of distress. The liner is also on a bearing of 315 from the ship. When you draw the diagram, you can mark down 45 degrees twice - one is between the 54 km and the 90 km, and one is between the 90km and the normal at that point. I am guessing that you have made an assumption that the ship is at a bearing of 315 from the lifeboat AFTER it has moved 54 km ...
--- End quote ---
Darren which year is this from so we can check our answer ?
J.Darren:
--- Quote from: Ari Ben Canaan on May 30, 2010, 01:39:09 pm ---Darren which year is this from so we can check our answer ?
--- End quote ---
Honeslty I have no idea LOL, perhaps you should ask the guy who posted the question :D
jellybeans:
--- Quote from: A@di on May 30, 2010, 12:45:59 pm ---check the above post.
ur answer is correct, whats the matter? ms trouble?
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:O guess so! thats what it says in the ms
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v
;)
--- Quote from: J.Darren on May 30, 2010, 01:41:44 pm ---Honeslty I have no idea LOL, perhaps you should ask the guy who posted the question :D
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AND I AM NOT A GUY.
... so is the MS right? D:
theharrovian:
I reckon J.Darren is correct, the problem is that everybody bar him has misinterpreted the question. The question states that "At 0600 hours the liner, which is 90 km from a lifeboat and on a bearing of 315 from the lifeboat...". All the other answers seem to assume that the liner is on a bearing of 315 from the lifeboat at 0730 hours, and not at 0600 hours.
cooldude, you've just made a computation error - everything is correct up to the bit when you somehow say that A is 133.75 km.
J.Darren:
--- Quote from: jellybeans on May 30, 2010, 02:13:00 pm ---:O guess so! thats what it says in the ms
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v
;)
AND I AM NOT A GUY.
... so is the MS right? D:
--- End quote ---
I do apologise :-[
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