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A pilot wants to fly on a bearing of 74.9 degrees. By flying due east, he find s that a 42-mph wind, blowin from the south, puts him on course. Find the airspeed and the groundspeed.
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|dw:1335749965540:dw|
try working with the tan function?
sin15.1 = 42 / x solve for x, yields: 161 mph for hypotenuse. This is the resultant net speed with reference to the ground (groundspeed). To find air speed we must find the remaining side of the triangle (the one horizontal with due east). From pythag 161^2 = 42^2 + a^2 a = 155 mph, which is the airspeed.
@Wolf77 - the OP did NO work here. do you think that is the intent of this site?
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