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    I have no idea what to do in this circle question. Could you also help me in the bearing question part c).

    Thanks.

    The answers are: from the circle one m=36;
    the bearing one is 308.
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    (Original post by ndk123)
    I have no idea what to do in this circle question. Could you also help me in the bearing question part c).

    Thanks.
    Hi! I moved this to the maths forum for you - you're more likely to get a good answer here
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    What will help you here is the the parallel lines tell you that the angle at P and O add up to 180. Also the angle at the opposite side of the circle from Q is twice the angle at the centre and also add to the angle at Q to give 180.

    Put these facts together to form an equation in m for the four angles in PQRS add ing up to 360
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    (Original post by nerak99)
    What will help you here is the the parallel lines tell you that the angle at P and O add up to 180. Also the angle at the opposite side of the circle from Q is twice the angle at the centre and also add to the angle at Q to give 180.

    Put these facts together to form an equation in m for the four angles in PQRS add ing up to 360
    Thanks. But i'm still not getting the correct answer.

    So:

    angle P=m
    angle O= 180-m
    angle Q= 180-2m
    Angle R=2m

    so
    m+180-m+180-2m+2m=360
    this gets all cancelled out.... So what am i supposed to do then?

    thanks.
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    For the angle corresponding to 2m you have to use the fact that the opposite angles of a cyclic Quad add to 180 (also the opposite angle to Q being half the angle at the centre O.) The opposite angle is half of the angle at O and is hence (180-m)/2.

    This gives you an angle at Q of 180-((180-m)/2)=(180+m)/2

    The working will be a good exercise but gives m=36

    Without using the fact that the opposite angles add to 180 and the you are effectively saying that 360=360 or in other words what you stated in
    m+180-m+180-2m+2m=360 is an identity not an equation.

    Because it would work for any quad like that.

    It is incorporating the opposite angle in the cyclic quad that constrains m to a value that works for a quad that is in a circle with vertices at the centre and circumference.

    The equation is
    m+180-m+(180+m)/2+2m=360 >>>> 2m+360-2m+180+m+4m=720 >>>>>>m=36

    Unless I have made an error.

    The identity-equation thing is a bit subtle for GCSE(IMHO)
 
 
 
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