What bearing and airspeed are required for a plane to fly 400 miles due north in 2.5 hours if the wind is blowing from a direction of 328 degrees at 11 mph?
(answer: 358 degrees; 170 mph) Show Calculations!
2007-07-01 13:23:01 · 3 answers · asked by daniel 1 in Science & Mathematics ➔ Mathematics
Draw the navigational triangle.
The long side (running north-south) is of length (400/2.5)mph.
The wind side is of length 11mph.
You can depict the wind side to show an included angle of (360º - 328º).
You know two side lengths (400 mph/2.5 and 11 mph) and the included (32º) angle. Go from there--you know, Laws of Cosines and Sines.
2007-07-01 13:46:06 · answer #1 · answered by Mark 6 · 0⤊ 2⤋
The wind vector is 11 mph in direction S32E.
The require ground speed is 400/2.5= 160 in a direction due North.
Thus using the law of cosines we get airplanes speed = sqrt ( 11^2+160^2 -2*11*160*cos(148) = 169.429 mph
to get plane's bearing we use law of sines:
sin x = 11sin148/169.429 = .0344044521
So x = arcsin .0344044521= = 1.97162 degrees which is a bearing of 360 - 1.97162 =358.02 degrees
2007-07-01 14:47:51 · answer #2 · answered by ironduke8159 7 · 1⤊ 1⤋
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2016-08-24 07:24:59 · answer #3 · answered by ? 4 · 0⤊ 0⤋