Algebra.Com's Answer #845440 by ikleyn(52787)![\"\"](\"/images/stars/stars-13.gif\")  
You can put this solution on YOUR website!
.
\n" );
document.write( "ABC is a triangle with ∠CAB=15∘ and ∠ABC=30∘.
\n" );
document.write( "If M is the midpoint of AB, find ∠ACM.
\n" );
document.write( "~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\r
\n" );
document.write( "
\n" );
document.write( "
\n" );
document.write( "\n" );
document.write( " The solution by Edwin is perfect and deserves admiration.\r
\n" );
document.write( "
\n" );
document.write( "\n" );
document.write( " My congratulations to Edwin with this great achievement.\r
\n" );
document.write( "
\n" );
document.write( "\n" );
document.write( " I came to bring a shorter solution.\r
\n" );
document.write( "
\n" );
document.write( "
\n" );
document.write( "\n" );
document.write( "\r\n" );
document.write( "The given part is shown in Figure 1 below.\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( " \r\n" );
document.write( "\r\n" );
document.write( " Figure 1.\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "Triangle ABC has the angles A= 15°, B= 30° and C= 180°-15°-30°= 135°.\r\n" );
document.write( "\r\n" );
document.write( "We want to find angle  .\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "Draw perpendicular CH from vertex C to the base AB (Figure 2).\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( " \r\n" );
document.write( "\r\n" );
document.write( " Figure 2.\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "First part of the solution is the same as that of Edwin' solution,\r\n" );
document.write( "\r\n" );
document.write( "and it shows that a =  , b =  .\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "The rest of the solution is different (very simple and totally geometrical).\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "Triangle BHC is a right-angled triangle with the acute angle B of 30°.\r\n" );
document.write( "Hence, the opposite leg CH is half of the hypotenuse BC\r\n" );
document.write( "\r\n" );
document.write( " CH =  =  , (1)\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "while its other leg BH is  times the hypotenuse BC\r\n" );
document.write( "\r\n" );
document.write( " BH =  =  .\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "Then the segment MH is the complement of BH to 1\r\n" );
document.write( "\r\n" );
document.write( " MH = 1 - =  = . (2)\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "Comparing expressions (1) and (2), we see that CH = MH.\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "Hence, right-angled triangle MHC is isosceles right-angled triangle,\r\n" );
document.write( "which implies that angle MCH is 45°.\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "Thus ∠ACM = 135° - 60° - 45° = 30°,\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "and the problem is solved completely.\r\n" );
document.write( " \r
\n" );
document.write( "
\n" );
document.write( "\n" );
document.write( " \n" );
document.write( " |