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help please maths

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.
(edited 7 years ago)
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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 :smile:
Reply 2
Avatar for nerak99
nerak99
13
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
Reply 3
Avatar for ndk123
ndk123
OP
14
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.
Reply 4
Avatar for nerak99
nerak99
13
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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