Turn on thread page Beta
    Offline

    2
    ReputationRep:
    (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?
    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
    Offline

    2
    ReputationRep:
    (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
    Offline

    6
    ReputationRep:
    (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?
    Offline

    2
    ReputationRep:
    (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
    Offline

    6
    ReputationRep:
    (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.
    Offline

    0
    ReputationRep:
    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
    Offline

    0
    ReputationRep:
    (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
    Offline

    1
    ReputationRep:
    (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.
    Offline

    3
    ReputationRep:
    whos doing ial C12 im scared the paper is 2 hours and 30 minutes wish i did gce
    Offline

    8
    ReputationRep:
    Erm... #AllNighterClub
    Offline

    2
    ReputationRep:
    (Original post by motheryucker)
    Erm... #AllNighterClub
    -high five-
    Offline

    1
    ReputationRep:
    (Original post by Skygon)
    Could anyone please help me with this question: Attachment 403237

    I would like to know where to start
    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?
    Offline

    16
    ReputationRep:
    (Original post by motheryucker)
    Erm... #AllNighterClub
    #EarlyMorningClub
    Offline

    2
    ReputationRep:
    (Original post by motheryucker)
    Erm... #AllNighterClub
    Yes!
    Offline

    3
    ReputationRep:
    Volume and sa for prism
    Offline

    0
    ReputationRep:
    #AllNighterClub #RedBULL #C1>C2 ANYDAYOFTHEWEEK
    Offline

    1
    ReputationRep:
    vol area of cross section x length
    sa total area of each face on prism
    Offline

    1
    ReputationRep:
    How you find area of an equilateral triangle

    Base times height /2 :hmmm:

    And also 0.5ab sinc
    Posted from TSR Mobile
    Offline

    2
    ReputationRep:
    OHk guyz gud luck to all of you out there
    Ace the exam!!!
    Offline

    12
    ReputationRep:
    Is the exam likely to be harder than last years ??? Omg feeling nervous now :-/

    Posted from TSR Mobile
 
 
 
Reply
Submit reply
Turn on thread page Beta
Updated: May 15, 2017

2,044

students online now

800,000+

Exam discussions

Find your exam discussion here

Poll
Should predicted grades be removed from the uni application process

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Write a reply...
Reply
Hide
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.