document.write( "Question 1375: obtuse isosolese triangle angles 120/30/30.equal legs 7.what is formula to determine third (longest leg/base)? \n" ); document.write( "
Algebra.Com's Answer #426 by khwang(438)\"\" \"About 
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\n" ); document.write( " Even we cannot have good diagram of triangle here, try to figure
\n" ); document.write( " out if angle BAC is 120 deg, both ABC & BCA are 30 deg, and AB=AC = 7.
\n" ); document.write( " A
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\n" ); document.write( " B D C\r
\n" ); document.write( "\n" ); document.write( " Let AD be the height of the base BC(note AD is perendicular to BC),
\n" ); document.write( " we ses that angle BAD = 120/2 =60, so BD/AB = sin 60 ,
\n" ); document.write( " BD = AB sin 60 = 7 * sqrt(3)/2 = 6.06
\n" ); document.write( " Since BD=CD,BC = 2 BD.
\n" ); document.write( " Hence,the longest leg BC = 2*BD = 12.12 \r
\n" ); document.write( "\n" ); document.write( " Of course, we can use the law of cosine directly, [Note: a=BC, b= AC, c=AB]
\n" ); document.write( " a^2 =b^2 + c^2 - 2bc cos A,
\n" ); document.write( " = 7^2 + 7^2 - 2*7^2 cos(120)
\n" ); document.write( " = 98(1 +0.5) [Note: cos 120 = -0.5]
\n" ); document.write( " = 147,
\n" ); document.write( " So, a = sqrt(147) = 12.12 (same answer)
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