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[i] An observer on top of a tower spots a fishing boat on a bearing of 340T at an angle of depression of 1.2 degrees at 1000 hours. At 1045 hours, the same boat is on a bearing of 070T at an angle of depression of 2.3 degrees. The observer is 60m above sea level.
I need to calculate the bearing, correct to one decimal place, from the trawler's 1000 hour's position and it's 1045 position.
Mathsux!
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Is it 132.5T? (Thats the bearing of the 1045 position relative to the 1000 position.)
That would be (090+θ)T, where
The actual height of the tower is not needed in the end. The angles of depression are much more important.
Last edited by JaneFairfax (2007-09-03 23:46:46)
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Answer: 132.45T degrees assuming the Earth is flat.
Notice it is a right triangle between observations.
Now 340T going back to observer is 340 - 180 = 160T
Now subtract 27.54 degrees from 160T to get the answer.
Why?
Because 1st observation is 2864.37 meters away, the long leg of the right triangle.
And 2nd observation is 1493.87 meters away.
Do inverse tangent to get the angles of the right triangle.
Last edited by John E. Franklin (2007-09-04 14:16:09)
igloo myrtilles fourmis
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