document.write( "Question 524976: The heading of a plane is 27.7 degrees NE, and its air speed is 255 mi.h. If the wind is blowing from the south with a velocity of 42.0 mi/h, find the actual direction of travel of the plane, and its ground speed. \n" ); document.write( "

Algebra.Com's Answer #347990 by Edwin McCravy(20055) $\"\"$ $\"About$

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The heading of a plane is 27.7 degrees NE, and its air speed is 255 mi.h. If the wind is blowing from the south with a velocity of 42.0 mi/h, find the actual direction of travel of the plane, and its ground speed.
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document.write( "The other tutor used the law of cosines and the parallelogram method.\r\n" );
document.write( "I will use the component method:\r\n" );
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document.write( "The red vector represents the wind's velocity.  Since the wind\r\n" );
document.write( "is FROM the SOUTH, it is blowing TOWARD the NORTH!  \r\n" );
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document.write( "The green vector represents the plane's velocity.\r\n" );
document.write( "Note that that the angle 27.7° is measured clockwise from\r\n" );
document.write( "the vertical (North), so the angle we will use is measured \r\n" );
document.write( "counterclockwise from the horizontal (East) calculated as \r\n" );
document.write( "90°-27.7° or 62.3°.\r\n" );
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document.write( "         x-components          y-components\r\n" );
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document.write( "Plane     255cos(90°)            255sin(90°)\r\n" );
document.write( "Wind       42cos(62.3°)           42sin(62.3°)\r\n" );
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document.write( "Sums:     19.52336592            292.1865323\r\n" );
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document.write( "|Resultant| =  = 292.8380635 mi/h\r\n" );
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document.write( "Angle (counterclockwise from East) = tan^-1( = 86.177°\r\n" );
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document.write( "To find the bearing clockwise from the North, we subtract from 90° and get 3.823°,\r\n" );
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document.write( "and the bearing is written as N 3.823° E.\r\n" );
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document.write( "I agree with the other tutor on the magnitude of the resultant, \r\n" );
document.write( "but we differ just a bit on the angle.\r\n" );
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document.write( "Edwin

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