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1. 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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2. (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
3. 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
4. (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.
5. 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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