document.write( "Question 1027374: A triangle is right- angled at B. D is the point on AC such that BD is perpendicular to AC. Let angle BAC=theta
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\n" ); document.write( "i. Given that 6AD+BC=5AC show that 6costheta+tantheta=5sectheta
\n" ); document.write( "ii. Deduce that 6sin^2theta-sintheta-1=0\r
\n" ); document.write( "\n" ); document.write( "I have already done part i, I just found what AD, BC AND AC are in terms of their trignometric rations. I then subbed them into the equation and received it.\r
\n" ); document.write( "\n" ); document.write( "I don't know how to do question ii
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Algebra.Com's Answer #642568 by mananth(16946)\"\" \"About 
You can put this solution on YOUR website!
6 cos \"theta\"+tan\"theta\" = 5sec\"theta\"\r
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\n" ); document.write( "\n" ); document.write( "6 cos \"theta\"+tan\"theta\" = 5/cos\"theta\"\r
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\n" ); document.write( "\n" ); document.write( "multiply by cos\"theta\"\r
\n" ); document.write( "\n" ); document.write( "6cos^2\"theta\" +tan\"theta\"*cos\"theta\" = 5\r
\n" ); document.write( "\n" ); document.write( "6cos^2\"theta\"+sin\"theta\" =5\r
\n" ); document.write( "\n" ); document.write( "6((1-sin^2\"theta\") +sin\"theta\"=5\r
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\n" ); document.write( "\n" ); document.write( "6- 6sin^2\"theta\" +sin\"theta\"=5\r
\n" ); document.write( "\n" ); document.write( "rearrange\r
\n" ); document.write( "\n" ); document.write( "6sin^2\"theta\" -sin\"theta\"-1=0\r
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