C1/C2 Questions Watch

creativebuzz
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(d) I managed to find the area of the sector quite easily, but I was stuck on how to do the area of the triangle because I wasn't sure how to find sine (because 1/2 x ab x sinC)



I managed to get this far:

(7-x)^2 = x^2 + (x+1)(x+1) - ((2(x)(x+1)(cos60))

But I wasn't sure how to expand/what to do with the cos60



I managed to get this far:

2^x = 5^x-1

xlg2=(x-1)lg5

But I wasn't sure what to do afterwards..
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Mr M
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(Original post by creativebuzz)
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9d) You got the angle in part c)?

2) You surely know the value of cos(60)?

5) Expand your bracket then collect x terms.
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creativebuzz
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(Original post by Mr M)
9d) You got the angle in part c)? Yup I got cosx = 7/25

2) You surely know the value of cos(60)? 1/2 omg of course! *facepalm*

5) Expand your bracket then collect x terms.
xlg2 = xlg5 - lg5

xlg2-xlg5 = -lg5

x(lg2-lg5) = -lg5

x(lg2/lg5) = -lg5

x(0.430..)= -0.69

x = 1.6255 Where have I gone wrong?
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Mr M
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(Original post by creativebuzz)
xlg2 = xlg5 - lg5

xlg2-xlg5 = -lg5

x(lg2-lg5) = -lg5

x(lg2/lg5) = -lg5

x(0.430..)= -0.69

x = 1.6255 Where have I gone wrong?
\displaystyle \log2 - \log 5 \neq \frac{\log 2}{\log 5}

You invented your own law of logarithms.
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creativebuzz
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(Original post by Mr M)
\displaystyle \log2 - \log 5 \neq \frac{\log 2}{\log 5}

You invented your own law of logarithms.
Why do I keep trying to invent new stuff? :P

But I've got the answer now, thank you very much!

Could you help me with question 10 (d) (the first picture in this thread)
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Mr M
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(Original post by creativebuzz)
Could you help me with question 10 (d) (the first picture in this thread)
It's pretty straightforward. What have you tried?
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creativebuzz
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(Original post by Mr M)
It's pretty straightforward. What have you tried?
I wasn't even sure when to start :/
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Mr M
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(Original post by creativebuzz)
I wasn't even sure when to start :/
What do you know about perpendicular lines?
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creativebuzz
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(Original post by Mr M)
What do you know about perpendicular lines?
Their gradient is the negative reciprocal to a parallel gradient
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Mr M
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(Original post by creativebuzz)
Their gradient is the negative reciprocal to a parallel gradient
So can you find the equation of the line that is perpendicular to L1 through C?
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creativebuzz
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(Original post by Mr M)
So can you find the equation of the line that is perpendicular to L1 through C?
Yup, I got y = x/2 + 3/2
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Mr M
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(Original post by creativebuzz)
Yup, I got y = x/2 + 3/2
Assuming that is right, if you solve it simultaneously with L1 then you will get the coordinates of D as the answer.
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creativebuzz
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(Original post by Mr M)
Assuming that is right, if you solve it simultaneously with L1 then you will get the coordinates of D as the answer.
Why am I solving it simultaneously with L1?
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Mr M
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(Original post by creativebuzz)
Why am I solving it simultaneously with L1?
Because the question says D is the point of intersection of L1 and the line you just found.
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creativebuzz
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(Original post by Mr M)
Because the question says D is the point of intersection of L1 and the line you just found.
I don't see where it says that
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Mr M
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(Original post by creativebuzz)
I don't see where it says that
"The point lies on L1" couldn't be much clearer. You seem to be tackling work that is too hard for you. Maybe you should do something easier and build back up to this?
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