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#2
(Original post by memeeee)
paper is here! anyone got any answers?
i got A, B, D, A, C, C, D, C, B, A for the multiple choice but it's very likely those are wrong.
paper is here! anyone got any answers?
i got A, B, D, A, C, C, D, C, B, A for the multiple choice but it's very likely those are wrong.
Also, http://www.wolframalpha.com/ is pretty useful for double checking some answers.
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(Original post by Blazy)
At a quick glance, you may want to take another look at Question A
Also, http://www.wolframalpha.com/ is pretty useful for double checking some answers.
At a quick glance, you may want to take another look at Question A
Also, http://www.wolframalpha.com/ is pretty useful for double checking some answers.
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#4
Does anybody have a quick solution for question 4, I have a solution but it is fairly longwinded
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#5
(Original post by mathphysics123)
Does anybody have a quick solution for question 4, I have a solution but it is fairly longwinded
Does anybody have a quick solution for question 4, I have a solution but it is fairly longwinded
Sketch solution:
Spoiler:
(a) area
is shown using 1/2 base x height, with the "height" being calculated using sin \alpha. Write 
for the area of ABC, then we can deduce the sine rule fom the fact that 
, then dividing by abc and taking the reciprocal of each fraction.
(b)
, so area BRP = 
, Similarly area CQP = 
and area ARQ = 
. Subtracting all 3 regions from the total area gives the desired result.
Show
(a) area
is shown using 1/2 base x height, with the "height" being calculated using sin \alpha. Write for the area of ABC, then we can deduce the sine rule fom the fact that , then dividing by abc and taking the reciprocal of each fraction. (b)
, so area BRP = , Similarly area CQP = and area ARQ = . Subtracting all 3 regions from the total area gives the desired result.
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#6
(Original post by DFranklin)
It's going to be hard to provide a solution without diagrams.
Sketch solution:
It's going to be hard to provide a solution without diagrams.
Sketch solution:
Spoiler:
(a) area is shown using 1/2 base x height, with the "height" being calculated using sin \alpha. Write for the area of ABC, then we can deduce the sine rule fom the fact that , then dividing by abc and taking the reciprocal of each fraction.
(b), so area BRP = , Similarly area CQP = and area ARQ = . Subtracting all 3 regions from the total area gives the desired result.
Show
(a) area
is shown using 1/2 base x height, with the "height" being calculated using sin \alpha. Write for the area of ABC, then we can deduce the sine rule fom the fact that , then dividing by abc and taking the reciprocal of each fraction. (b)
, so area BRP = , Similarly area CQP = and area ARQ = . Subtracting all 3 regions from the total area gives the desired result.
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#7
(Original post by Math1791)
The solution to part b only makes sense if none of the angles are obtuse, and the question does not mention that alpha,beta,delta are less than equal to 90 degrees.
The solution to part b only makes sense if none of the angles are obtuse, and the question does not mention that alpha,beta,delta are less than equal to 90 degrees.
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#8
(Original post by mathphysics123)
Does anybody have a quick solution for question 4, I have a solution but it is fairly longwinded
Does anybody have a quick solution for question 4, I have a solution but it is fairly longwinded
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#9
(Original post by DFranklin)
I'll give you 10:1 it's the expected solution, however.
I'll give you 10:1 it's the expected solution, however.
PS I haven't thought about this in detail, but I imagine the question still makes sense for an obtuse triangle if the result is appreciated in terms of signed areas.
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#10
(Original post by Math1791)
The solution to part b only makes sense if none of the angles are obtuse, and the question does not mention that alpha,beta,delta are less than equal to 90 degrees.
The solution to part b only makes sense if none of the angles are obtuse, and the question does not mention that alpha,beta,delta are less than equal to 90 degrees.
(Original post by RichE)
Yes I recall this being noted after the fact.
Yes I recall this being noted after the fact.
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#11
(Original post by DFranklin)
@Math1791: Your ability to spot corner cases in MAT questions is really impressive (I'm not being sarcastic), but I really hope you're not going to let it stop you finding the expected solutions when it comes to the actual exam. When sitting an exam, there are times when you have to go with "there's a possible complication, but given the flow of the question and how different a solution dealing with the complication would be, I'm just going to ignore it" (possibly mention it if you're concerned).
@Math1791: Your ability to spot corner cases in MAT questions is really impressive (I'm not being sarcastic), but I really hope you're not going to let it stop you finding the expected solutions when it comes to the actual exam. When sitting an exam, there are times when you have to go with "there's a possible complication, but given the flow of the question and how different a solution dealing with the complication would be, I'm just going to ignore it" (possibly mention it if you're concerned).
I will keep it in mind and try my best to follow it.
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#12
Can someone help me with question 3 on this paper.
Part A is fine, but from part B on, I am unsure.
I had to form 4 equations using various values of T and then create a matrix. Using the inverse matrix I found the 4 coefficients and found the cubic.
But how are you actually meant to do this question.
Part A is fine, but from part B on, I am unsure.
I had to form 4 equations using various values of T and then create a matrix. Using the inverse matrix I found the 4 coefficients and found the cubic.
But how are you actually meant to do this question.
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