document.write( "Question 1207590: In triangle $PQR,$ let $X$ be the intersection of the angle bisector of $\angle P$ with side $QR$, and let $Y$ be the foot of the perpendicular from $X$ to side $PR$. If $PQ = 10,$ $QR = 10,$ and $PR = 12,$ then compute the length of $XY$. \n" ); document.write( "

Algebra.Com's Answer #845579 by Edwin McCravy(20056) $\"\"$ $\"About$

You can put this solution on YOUR website!
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document.write( "In triangle PQR, let X be the intersection of the angle bisector of angle P \r\n" );
document.write( "with side QR, and let Y be the foot of the perpendicular from X to side PR.\r\n" );
document.write( "If PQ = 10, QR = 10, and PR = 12, then compute the length of XY.\r\n" );
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document.write( "Triangle PQR is isosceles, so let the base angles at P and R be 2θ\r\n" );
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document.write( "Angle bisector PX divides QR=10 into the ratio of PQ:PR = 10:12 = 5:6,\r\n" );
document.write( "so it's easy to show that QX=50/11 and XR=60/11. [Note that 50/11+60/11=10].\r\n" );
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document.write( " because it is an exterior angle of triangle PRX.  \r\n" );
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document.write( "Using the law of sines on triangle PXQ\r\n" );
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document.write( "Multiply both sides by 11/10\r\n" );
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document.write( "We look up the formula \r\n" );
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document.write( "Multiply through by sin(θ):\r\n" );
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document.write( " => \r\n" );
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document.write( "Now look at right triangle XYR\r\n" );
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document.write( "Edwin

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