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An aeroplane leaves airport flies due north for 2hrs at 500km/hr it then flies 450km on a bearing of53 how far is the aeroplane from the airport
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Draw a diagram, then use the law of cosines to find the distance.
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An aeroplane leaves an airport, flies due north for 2hrs at 500km/hr.it then flies 450km on a bearing 053.how far is the plane
Top answer:
the plane -- starting at (0,0) -- ends up at (x,y) = (450 sin53°,2*500 + 450 cos53°) and, as
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an aeroplane leaves an airport, flies due north for 2hrs at 500km/h. It then flies on a bearing of 053 degree at 300km/h for
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use the law of cosines. The distance z is z^2 = 1000^2 + 450^2 - 2*1000*450 cos127°
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an aeroplane leaves an airport,flies due north for two hours at 500km/hr.it then flies 450km on a bearing 053.how far is the
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using the law of cosines, the distance z is z^2 = 1000^2 + 450^2 - 2*1000*450 cos127° If the plane
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an aeroplane leaves an airport,flies due north for two hours at 500km/hr.it then flies 450km on a bearing 053.how far is the
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1st leg goes (0,2*500) = (0,1000) 2nd leg goes (450sin53°,450cos53°) = (359,271) so, the plane
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I don't mind helping to find answers, but I resent having to provide the questions as well.
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(1). An Aeroplane leaves an airport, flies due North for 2 hrs at 500km/h. It then flies on a bearing of 053 degree at 300km/h
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(1) you fly on a <u>heading</u>, not a bearing. The law of cosines is the way to find this value.
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North 500 + 300 cos 60 = 650 East 300 sin 60 = 260 tan bearing from origin = East/North =260/650 =
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