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19 February, 13:35

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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  1. 19 February, 16:05
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    The bearing shows the angle from north to the line (c), you want to see the angle inside the triangle so that will be 90-[bearing].

    a=42 (42 mph wind blowing north)

    A = 90-74.9

    A = 15.1 degrees

    Ground speed is the speed the plane is going including the wind.

    ground speed = c

    Airspeed = b

    You have the angle, and you have the "Opposite" and want to find the "Hypotenuse".

    SOH CAH TOA

    Sin x = Opp/Hypot

    Sin (15.1) = 42/c

    c = 42/Sin (15.1)

    c = 161.53

    Ground speed of the plane is 161.53 mph

    Airspeed = b

    Tan x = Opp/Adj

    Tan (15.1) = 42/b

    b = 42/Tan (15.1)

    b = 161.53

    Airspeed = 161.53 mph

    (so the answer is ground speed of 161.53mph)
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