Edexcel C2 20th May 2015 Official Thread

Part d on this question confuses me. I've drawn a diagram, but can't get my head around how they got the angle ARQ. I drew a diagram from my understanding of the situation. Is there anything wrong with it?

You're diagram is incorrect so it's hard for you to understand, draw it again with some perspective with regards to the points P and Q as they do not lie on the same line like you have drawn

Think about perpendicular bisected of PR as the shortest distance then use Pythagoras, but you won't be able to do this without a diagram that is correct

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Reply 1701

johnhick98

2

Original post by AdamRazaa

What topics are best to keep fresh in my mind for smaller non long winded questions?

Geometric sequences and trigonometrical identities are worth many marks but not too complictaed

Reply 1702

dominicwild

6

Original post by domcandrews

You're diagram is incorrect so it's hard for you to understand, draw it again with some perspective with regards to the points P and Q as they do not lie on the same line like you have drawn

Think about perpendicular bisected of PR as the shortest distance then use Pythagoras, but you won't be able to do this without a diagram that is correct

Posted from TSR Mobile

I don't understand how P and Q don't lie on the same line. As the question states that, a circle with center A, goes through points P and Q. Therefore they lie on the circle and PQ is the diameter. The center point, is the mid point of the two, so they line on the same line?

Reply 1703

domcandrews

2

Original post by dominicwild

I don't understand how P and Q don't lie on the same line. As the question states that, a circle with center A, goes through points P and Q. Therefore they lie on the circle and PQ is the diameter. The center point, is the mid point of the two, so they line on the same line?

ok yeah they lie on the sameline, but its easier to picture it if you points in perspective i.e rather than how youve done it, maybe its just me i dunno

Reply 1704

dominicwild

6

Original post by domcandrews

ok yeah they lie on the sameline, but its easier to picture it if you points in perspective i.e rather than how youve done it, maybe its just me i dunno

Well, I can't imagine any other way to draw it really.

Reply 1705

Lukemarks1997

Use the cosine rule to figure out the angle as you have all of the sides of a triangle. a^2 = b^2 + c^2 - 2bc cosA where A is your angle

Reply 1706

Lukemarks1997

Original post by Lukemarks1997

Use the cosine rule to figure out the angle as you have all of the sides of a triangle. a^2 = b^2 + c^2 - 2bc cosA where A is your angle

Actually I think in the mark scheme they used sine rule but both should work I think. Dunno

Reply 1707

Skygon

1

Original post by dominicwild

Part d on this question confuses me. I've drawn a diagram, but can't get my head around how they got the angle ARQ. I drew a diagram from my understanding of the situation. Is there anything wrong with it?

Consider the midpoint of PR to be M. If you find the angle of APM, it's the same as ARQ.

5 root 5 / 15 = Sin theta

Solve to find theta.

It is also worth noting, if you find the angle subtended at the centre, by the chord, you can avail of the cosine rule to solve it.

(edited 8 years ago)

Reply 1708

SOREN_TORETTO

3

whos doing ial C12 im scared :frown:

the paper is 2 hours and 30 minutes wish i did gce

Reply 1709

motheryucker

8

Erm... #AllNighterClub

Reply 1710

marco cb

2

Original post by motheryucker

Erm... #AllNighterClub

-high five-

Reply 1711

Hsakbo

1

Original post by Skygon

Could anyone please help me with this question: Screen Shot 2015-05-20 at 3.50.53 AM.png

I would like to know where to start :s-smilie:

The ways is to draw a chord OB and OC and come to the conclusion that triangle Odc and OAB are equilateral and symmetric. Use the area of the sector and segment equations and minus the symmetric parts from the main semi.

Am I too late?