document.write( "Question 1138788: How do I find the largest possible area of an isosceles triangle if the length of each of the two equal sides is 10 m? \n" ); document.write( "
Algebra.Com's Answer #756575 by josmiceli(19441)\"\" \"About 
You can put this solution on YOUR website!
The height bisects the angle between 10m sides
\n" ); document.write( "Call these angles both \"+theta+\"
\n" ); document.write( "The height is \"+10%2Acos%28+theta+%29+\"
\n" ); document.write( "one-half of the base is \"+10%2Asin%28+theta+%29+\"
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\n" ); document.write( "The area is:
\n" ); document.write( "\"+A+=+%281%2F2%29%2A%28+10%2Asin%28theta%29+%29%2A10%2Acos%28theta%29+%29+\"
\n" ); document.write( "\"+A+=+50%2Asin%28theta%29%2Acos%28theta%29+\"
\n" ); document.write( "Use trig identity
\n" ); document.write( "\"+A+=+50%2Asin%28+2theta+%29\"
\n" ); document.write( "The rate of change is the cos, so
\n" ); document.write( "\"+A%5Bprime%5D+=+50%2Acos%28+2theta+%29+\"
\n" ); document.write( "Max is where \"+A%5Bprime%5D+=+0+\"
\n" ); document.write( "\"+50%2Acos%28+2theta+%29+=+0+\"
\n" ); document.write( "\"+2theta+=+pi%2F2+\"
\n" ); document.write( "\"+A%5Bmax%5D+=+50%2Asin%28pi%2F2%29+\"
\n" ); document.write( "\"+A%5Bmax%5D+=+50+\" m2
\n" ); document.write( "------------------------------
\n" ); document.write( "The angles are \"+pi%2F2+\", \"+pi%2F4+\", \"+pi%2F4+\"
\n" ); document.write( "-------------------------------------------------------
\n" ); document.write( "Get a 2nd opinion if needed.
\n" ); document.write( "Check my math, too\r
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