document.write( "Question 1192620: In triangle ABC, (bsin C) (bcos C + ccos B) =42\r
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Algebra.Com's Answer #824514 by MathLover1(20850)\"\" \"About 
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\r
\n" ); document.write( "\n" ); document.write( "given: \"%28b%2Asin%28+C%29%29+%28b%2Acos+%28C%29+%2B+c%2Acos%28+B%29%29+=42\"\r
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\n" ); document.write( "\n" ); document.write( "\"b%2Acos%28C%29+%2B+c%2Acos%28B%29\" is the length of the side \"a\" of the triangle ABC\r
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\n" ); document.write( "\n" ); document.write( "use the law of cosine to prove it:\r
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\n" ); document.write( "\n" ); document.write( "\"cos%28C%29+=++%28a%5E2+%2B+b%5E2+-+c%5E2%29%2F%282ab%29\"\r
\n" ); document.write( "\n" ); document.write( "\"cos%28B%29+=++%28c%5E2+%2B+a%5E2+-+b%5E2%29%2F%282ca%29\"\r
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\n" ); document.write( "\n" ); document.write( "then we have\r
\n" ); document.write( "\n" ); document.write( "\"b%2Acos+%28C%29+%2B+c%2Acos%28+B%29\"\r
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\n" ); document.write( "\n" ); document.write( "=\"%28a%5E2+%2B+b%5E2+-+c%5E2%2Bc%5E2+%2B+a%5E2+-+b%5E2%29%2F%282a%29\"\r
\n" ); document.write( "\n" ); document.write( "=\"%282a%5E2%29%2F%282a%29\"\r
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\n" ); document.write( "\n" ); document.write( "then we have\r
\n" ); document.write( "\n" ); document.write( "\"b%2Asin%28C%29\" is the length of the altitude drawn from vertex A to the side \"a\", and the area of triangle is:\r
\n" ); document.write( "\n" ); document.write( "\"area=%281%2F2%29ab%2Asin%28+C%29+\"\r
\n" ); document.write( "\n" ); document.write( " \"area=%281%2F2%2942\"\r
\n" ); document.write( "\n" ); document.write( "\"area=21\" sq. units\r
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