document.write( "Question 1070579: An aircraft fliesrounda triangular course. The first leg is 200km on a bearing of 115^0 and the second leg is 150km on a bearing of 230^0. How long is the third leg of the course and what bearing must the aircraft fly? \n" ); document.write( "
Algebra.Com's Answer #685633 by Boreal(15235)\"\" \"About 
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Draw this.
\n" ); document.write( "The SE corner angle is 115 degrees, and we will call that C.
\n" ); document.write( "the third leg will be c
\n" ); document.write( "c^2=a^2+b^2-2ab cos C
\n" ); document.write( "=40000+22500-2(200)(150)cos 115
\n" ); document.write( "=40000+22500+25357, the cosine of 115 being negative
\n" ); document.write( "c^2=87857.10
\n" ); document.write( "c=296.41 km, consistent with continuing to move away from the starting point on the second part.
\n" ); document.write( "For the angles, use the law of sines
\n" ); document.write( "sin 115/296.41=sin x/150
\n" ); document.write( "sin 115*150/296.41=0.4586
\n" ); document.write( "arc sin gives 27.3 degrees, this being the angle between the first and third legs.
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\n" ); document.write( "\n" ); document.write( "last one is sin y/200=sin 115/296.41
\n" ); document.write( "sin y=200*sin 115/296.41
\n" ); document.write( "sin y=0.6115
\n" ); document.write( "arc sin gives 37.7 degrees
\n" ); document.write( "The angles add to 180 degrees.
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\n" ); document.write( "\n" ); document.write( "The angle from the second leg to the final leg is 37.7 degrees
\n" ); document.write( "The reciprocal bearing is 50 degrees, and the angle is taken to the left of that or northward, so that the bearing is 12.3 degrees.
\n" ); document.write( "12.3 degree bearing for 296.41 km\r
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