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OCR FSMQ Additional Maths 6th June 2016 Official Thread

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That was a walk in the park, should have made it at least a little harder

Vanilla Cupcake

Original post by deardood

Was it not 97 something degrees as you had to make b and c smaller by 0.5 cm and a larger by 0.5cm. To get largest angle

i got 97.something too, idrk what i did but played around with some numbers for a bit and that's the highest i got

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Original post by hebeb

Yes this is what I got. Can't remember the exact figure but it was ~97 and I did what you did.

Just such a hard test I couldn't be sure on anything...

Original post by master123prick

you're a ****

well said matee

Doesnt't Mr M usually make a markscheme after this one?

Original post by toogoodatmaths

That was a walk in the park, should have made it at least a little harder

you're an absolute banging prick

Original post by master123prick

you're an absolute banging prick

Thats not very nice

username1828143

Original post by master123prick

you're an absolute banging prick

take no notice, they're being an attention-seeking troll.

OP

Original post by Irf00

Doesnt't Mr M usually make a markscheme after this one?

Not sure. @Mr M?

Original post by McSingh

well said matee

mcsingh cheers and did you get 28372837198981273123 degrees for the quadrilateral horizontal perpendicular integration line question

someone plz tell me the grade boundaries will be like 50 for an A

Original post by jadeee18x

What did everyone get for the maximum value of the inequality question?

45?

Original post by master123prick

mcsingh cheers and did you get 28372837198981273123 degrees for the quadrilateral horizontal perpendicular integration line question

naa i got 28372837198981273124

sorry mate, probably a trivial mistake

Vanilla Cupcake

Original post by ihatehannah

45?

I got this too 🙌🙌🙌

Original post by deardood

Was it not 97 something degrees as you had to make b and c smaller by 0.5 cm and a larger by 0.5cm. To get largest angle

That's what I did and I got 97.7 degrees

Original post by toogoodatmaths

Thats not very nice

shut the fck up you ****

Original post by McSingh

shut the fck up you ****

lol

Hi!
I was just wondering, does any one know if an unofficial mark scheme for today's paper has been created yet?
Thanks

Original post by SunilP_123

It's got nothing to do with lower bounds. Its just the upper bound of the side opposite to the angle. Other sides don't natter.

I did this and got 97.7...From the cos graph, between 0 and 180 degrees, the lower the cosine value, the higher the angle. Therefore, to get the maximum angle, you want to make your cosine value as low as possibleHere you could, so because (between 90 and 180) the angle would be larger if the cosine value was more negativecos B = (BC^2 + AB^2 - AC^2)/ (2BC x AB)Since the - AC part is negative, you want to use the maximum value of AC (20.5).The earlier part of the question was a hint to use 11.5.Then if you try 15.5 and then 16.5 for the remaining side, it turns out that 15.5 gives the larger angle, which was 97.7 (3s.f.)

this is how i did it

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