The Student Room Group

UNOFFICIAL MARKSCHEME Edexcel Maths Calculator paper 09/06/2016

Scroll to see replies

Original post by Certified
no it wasn't because 56 was corresponding, then its angles on a straight line..


I got 55
24) A) -1/3x+4

B0 2x/(x+1)(x-1)

5) 0.09, 0.36

13) -1,1,-1

11) 42.28
12) 144
Reply 262
Original post by rayestar
u mean pentagon righttt? :d


yessss is it 144 omggg yas
Original post by Jf1234
What if we put 9/20 instead of 0.45 for 1b?


I did this it will still give u a mark
What did everyone get for the pentagon question?
Fast baller..star sigining for the paki internetional team also porn all the way and btw 1+1=69
What is the predicted grade boundaries for an A* for this year? And have grade boundaries ever been more than 170/200 for an A*?
Original post by LilyKitty
I don't think so because the angle has to be between the two sides (7 &8) for the formula of triangle area to work and x was another angle so you had to do cosine to find another side and then sine rule too before you could get x. Not sure though. xx


What? I got 80.4 but used cousins twice. Is that all right?
Original post by Rajive
I got 40.0?


Yeah that's what I got too :smile: , seems to be a few different answers people got though so not sure if it's right
Original post by 34908seikj
Wasn't the last question like 1/2 * 7 * 8 Sin(X) = 18


56Sin(x) = 36

Sin^-1(36/56) = X

x = 40

???

I think I did it wrong then.


I got the same and I thought it's right but apparently not since everyone else got 80.4 or 80.1


Posted from TSR Mobile
GUYS WHAT DO YOU THINK THE GRADE BOUNDARIES WILL BE I REALLY WANT AN A BUT I THINK I GOT AROUND 120-130 OUT OF 200?!?!?! IS it possible?
Original post by ellamoon_22
This is correct. I forgot to do the inverse but i shall at least get 3/4marks i think


This is correct for the angle at B, but the question asked for the angle at A.
80.1 is also ok for the last question right?
Reply 273
3.8 is right
did u guys get 84638 bars ?
last question-
find angle bac

first find angle abc
18=0.5x7x8xsinabc
18=28sinabc
18/28= sinabc =0.642...
sin-1(0.642...)=abc=40.0052...

then find length AC
ac^2=7^2+8^2-3x7x8cos40.00
ac^2=27.209...
ac=5.216...

then find angle bac
18=0.5 x5.216...x7xsinbac
18/18.25...=sinbac=0.985...
sin-1(0.98...)=bac=80.3755...
=80.4
Original post by 34908seikj
Wasn't the last question like 1/2 * 7 * 8 Sin(X) = 18


56Sin(x) = 36

Sin^-1(36/56) = X

x = 40

???

I think I did it wrong then.



you had to find the area, then use the cosine rule then the sine rule to get 80.4
Original post by samiuls
yep its correct


W h a t!! !! I got 4! ! Time to cry salty tears.
Reply 278
I got 40.0 too...
Working back from 1/2absinC
Original post by hannah2360
What did everyone get for the pentagon question?


144 degrees

Quick Reply