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WJEC C2 January 2014 [17/1/14]

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Reply 40
Original post by sunflower95
Maybe we're right! :wink: ahha


I did something similar, you couldn't factorise them directly which proved they didn't


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Reply 41
what did you get for the equation of tangent how did you find the gradient
Reply 42
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Original post by sunflower95
Maybe we're right! :wink: ahha


Well it looks like I was wrong then!

What about the 2 triangles with the cosine rule? I struggled there.
Reply 43
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Original post by hii21
what did you get for the equation of tangent how did you find the gradient


I had y + 7 = 3/4(x-6) Which I think is wrong.

Diff in y / Diff in x for the gradient: -4-(-7) / 6-2

I don't think it's right though.. because that gradient needed to be flipped or something..
(edited 10 years ago)
Reply 44
Original post by hii21
what did you get for the equation of tangent how did you find the gradient


The gradient of the radius I that point was -3/4 so you flipped it for the tangent 4/3. I got something like 4x-3y- a number I can't remember


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Reply 45
Original post by HGN
I had y + 7 = 3/4(x-6) Which I think is wrong.

Diff in y / Diff in x for the gradient: -4-(-7) / 6-2

I don't think it's right though..


I did the same and so did everyone who did the exam with me but I'm sure it was 4/3 as the radius gradient was -3/4
It should be -7-(-4) over 6-2 as the points were (6,-7) and (2,-4) ?
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(edited 10 years ago)
Original post by mabli
I did the same and so did everyone who did the exam with me but I'm sure it was 4/3 as the radius gradient was -3/4


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Yes I had the 4/3 one!


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Original post by sunflower95
Maybe we're right! :wink: ahha


I reckon we are :tongue:


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Reply 48
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What did the Cos^2theta one come out as? 2(a) I think it was.
Reply 49
It was one of the worst papers I have ever sat. was it just me or was 6.b so frustrating?!
Original post by HGN
Well it looks like I was wrong then!

What about the 2 triangles with the cosine rule? I struggled there.


yeah i struggled on that one.. i came up with the two cos answers, then on the second part of the question it was like find x.. i completely forgot how to do the rearrangements after i put something like.. x^2+16/6x + x^2+35/2x =180.. ( i know thats not what the answer to part one was but something similar)
Original post by dm1818
It was one of the worst papers I have ever sat. was it just me or was 6.b so frustrating?!

YES! 6b was horrible.. i thought i could do it then i realised that we have cos, but area of a triangle is 1/2absinc so i didnt know how to turn cos to sin.. i tried with cos^2x+sin^2x=1 but i failed bad ahhah

Opps I meant 5b sorry
(edited 10 years ago)
Reply 52
the first answer was 3.502, did you get 4 possible solutions for 2a?
(edited 10 years ago)
Original post by sunflower95
YES! 6b was horrible.. i thought i could do it then i realised that we have cos, but area of a triangle is 1/2absinc so i didnt know how to turn cos to sin.. i tried with cos^2x+sin^2x=1 but i failed bad ahhah


I used the values of the bigger triangle (the two triangles formed a bigger triangle with all the sides given) and used one of the equations worked out in 6a to get cosB.

CosB came out as like 1/8 and then the angle B came out then as about 83 degrees.

I then subbed it into the halfABSinC equation which (I don't know if this is right) then rearranged to 1/2ACSinB and came out with an answer.

Can I rearrange the area equation like that?


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Original post by sunflower95
YES! 6b was horrible.. i thought i could do it then i realised that we have cos, but area of a triangle is 1/2absinc so i didnt know how to turn cos to sin.. i tried with cos^2x+sin^2x=1 but i failed bad ahhah


You could have cos to the -1 it then sin it
Reply 55
does anyone have the papers? can you upload it please
Reply 56
Only welsh copy sorry


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Reply 57
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Original post by mabli
Only welsh copy sorry


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Better than nothing!
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Reply 59
Did someone else end up with an algebraic fraction to solve in 6. b)?

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