W00t! I did question 10... (1 Viewer)

acmilan

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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
 

smallcattle

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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

arhh.... damn it.....i forgot to solve cos A

i was like cancelling, dividng, multiplying and stuff, and i thought wtf, how do i cancel cosA to get the fina answer.... :vcross:
 

euphoric

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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)
are u seirous?
thats soo easy! and i couldnt figure it out in the exam :vcross:
blha~ i only left that question out. besides that question ten was good!
 

skankit

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superbird said:
10B iii was good. all u had to do was differentiate it twice and then sub in the x value
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 minimum

i got the last part, but i didnt know how to factorise it :eek: thank god for them giving it in the question!!!]

i couldnt do the last part of 8b) i ended up with like 17 1/3 or something, i did all this hard core tricky intergration then i got and realised how easy it is :mad:

stupid maths

and i coudlnt for the life of me do question 5b iv.. i kept getting v=0... in the end i got something random which im positive wasnt right
 

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