smallcattle
Member
I couldnt get the very last question of Q10b
can someone explain??
can someone explain??
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and i quote: "All questions are of equal value"+GriM ReApeR+ said:hehehe...but failed all the other questions though![]()
acmilan1987 said:Using the z equation you found in (ii), sub in the x value of (iii). Now after simplification you will find z = root(bc - bccosA) and then you must find cosA using the large triangle ABC using the cos rule, sub the value for cosA into the equation of z and after some manipulations you will get what you need
are u seirous?acmilan1987 said:Since Area(AST)=Area(BCST) then
Area(ABC)=2Area(AST)
1/2(bcsinA) = 2.1/2(xysinA)
1/2(bcsinA) = xysinA
hence xy = 1/2(bc)
you didnt need to differentiate it twice, only once. you just leave it as z2 and differentiate that, and then let it equal 0 to find the minimumsuperbird said:10B iii was good. all u had to do was differentiate it twice and then sub in the x value