document.write( "Question 64783:
\r\n" ); document.write( "In a triangle we have ÐA=60°, ÐB=80° and ÐC=40°\r\n" ); document.write( "we draw median, angle bisector and altitude from vertex A to \r\n" ); document.write( "opposite side(BC). what are the value of all angles that \r\n" ); document.write( "are constructed on vertex A.
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Algebra.Com's Answer #46372 by Edwin McCravy(20060) $\"\"$ $\"About$

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document.write( "In a triangle we have ÐA=60°, ÐB=80° and ÐC=40°\r\n" );
document.write( "we draw median, angle bisector and altitude from vertex A to \r\n" );
document.write( "opposite side(BC).  what are the value of all angles that \r\n" );
document.write( "are constructed on vertex A.\r\n" );
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document.write( "Let D be the midpoint of BC. Then AD is the median. \r\n" );
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document.write( "AB/sin40° = BC/sin60° = AC/sin80°\r\n" );
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document.write( "To make things easier, let\r\n" );
document.write( "p = sin40° = .6427876097\r\n" );
document.write( "q = sin60° = .8660254038 \r\n" );
document.write( "r = sin80° = .984897753\r\n" );
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document.write( "AB/p = BC/q = AC/r\r\n" );
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document.write( "AB = BC·p/q\r\n" );
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document.write( "AC = BC·r/q\r\n" );
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document.write( "We calculate median AD using the law of cosines on DACD\r\n" );
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document.write( "AD² = AC² + CD² - 2·AC·CD·cos40°\r\n" );
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document.write( "substitute CD = BC/2 and AC = BC·r/q\r\n" );
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document.write( "AD² = BC²r²/q² + BC²/4 - 2·BC·r/q·BC/2·cos40°\r\n" );
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document.write( "AD² = BC²r²/q² + BC²/4 - BC²r·cos40°/q\r\n" );
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document.write( "Factor out BC² on the right:\r\n" );
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document.write( "AD² = BC²(r²/q² + 1/4 - r·cos40°/q)\r\n" );
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document.write( "AD = BC·Ör²/q² + 1/4 - r·cos40°/q\r\n" );
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document.write( "Let k = Ör²/q² + 1/4 - r·cos40°/q = .8197650971\r\n" );
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document.write( "So\r\n" );
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document.write( "AD = BC·k\r\n" );
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document.write( "By the law of sines in triangle ACD\r\n" );
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document.write( "CD/sin(ÐCAD) = AD/sin40²\r\n" );
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document.write( "CD/sin(ÐCAD) = AD/p\r\n" );
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document.write( "sin(ÐCAD) = CD·p/AD\r\n" );
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document.write( "Substitute CD = BC/2 and AD = BC·k\r\n" );
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document.write( "sin(ÐCAD) = (BC/2)·p/(BC·k)\r\n" );
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document.write( "the BC's cancel and we have\r\n" );
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document.write( "sin(ÐCAD) = p/(2k) = .6427876097/(2·.8187650971)\r\n" );
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document.write( "     = .3920559755\r\n" );
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document.write( "ÐCAD = 23.08248883°\r\n" );
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document.write( "---------------------\r\n" );
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document.write( "Let E be the point where the bisector of ÐA\r\n" );
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document.write( "Then ÐCAE = 30° because that's half of 60°\r\n" );
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document.write( "Draw altitude AF perpendicular to BC. DCAF is\r\n" );
document.write( "a right triangle and so ÐCAF is complementary\r\n" );
document.write( "to ÐC, which is 40° and so\r\n" );
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document.write( "ÐCAF = 50° \r\n" );
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document.write( "Now you have three of the angles at A.\r\n" );
document.write( "You can easily find any of the other angles\r\n" );
document.write( "at A by simple addition and subtraction.\r\n" );
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document.write( "Edwin

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