document.write( "Question 776252: find the area of the triangle whose longest and shortest sides are 62 cm and 31 cm respectively and whose biggest angle is 140 \n" ); document.write( "

Algebra.Com's Answer #473462 by josgarithmetic(39620) $\"\"$ $\"About$

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Two shorter sides are 31 and d in length and longest side is 62, opposite of the given angle. The given angle is between the 31 and d lengths.\r
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\n" ); document.write( "\n" ); document.write( "Law of Cosines:
\n" ); document.write( " $\"d%5E2%2B31%5E2-2%2A3%2Ad%2Acos%28140%29=62%5E2\"$
\n" ); document.write( " $\"d%5E2%2B31%5E2%2B2%2A3%2Ad%2Acos%2840%29=62%5E2\"$
\n" ); document.write( " $\"d%5E2%2B6%2Acos%2840%29%2Ad%2B31%5E2-62%5E2=0\"$ ---------- a quadratic equation in d. \r
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\n" ); document.write( "\n" ); document.write( "Finding value for d just requires computation through the general solution to the quadratic formula. After than, you can get the height of the triangle according to $\"d%2Acos%2840%29\"$ . Area of the triangle would be $\"%281%2F2%2931%2Ad%2Acos%2840%29\"$ . That just uses area as $\"%281%2F2%29%2Abase%2Aheight\"$ . \n" ); document.write( "

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