Precal Question. Please help!?

An airplane is heading due south with an airspeed of 192 mph. A wind from a direction of 53.0 degrees is blowing at 13.0mph. Find the bearing of the plane.

2 Answers

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  • 2 years ago
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    Presumably "from a direction of 53.0 degrees" means the wind is blowing approximately towards the SW or WSW.

    So add the airspeed, -192 j, to the wind speed, which is

    13.0 (i cos(233) + j sin(233)) = -7.8236 i - 10.3823 j.

    The resultant speed is -7.8236 mph i - 202.3823 j,

    a direction a little bit west of south.

    Arctan(7.8236/202.3823) = 2.21 degrees, so the bearing of the plane is

    182.21 degrees.

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  • david
    Lv 7
    2 years ago

    13 sin 53 = 10.4 .... 13 cos 53 = 7.8

    192 - 10.4 = 181.6

    tan A = 181.6 / 7.8 .. A = 87.5 deg

    182.5 deg. from true north

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