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C2 - Circles - Help

Been stuck on part (b) of this circles question, don't quite understand it:



I get that if we split it into two, we get the angle to be pi/6, and the radius of the circle to be 6-r, but don't know what to do from there.
(edited 8 years ago)
Reply 1
I think you could use sohcahtoa (sin=o/h), and do sin(pi/6) = r/6-r
So if you do sin(pi/6) you get 1/2, so 1/2 = r/6-r
6-r = 2r and so on until r=3
(edited 8 years ago)
Original post by Reety
Been stuck on part (b) of this circles question, don't quite understand it:



I get that if we split it into two, we get the angle to be pi/6, and the radius of the circle to be 6-r, but don't know what to do from there.


Consider a radii of the circle which is perpendicular to the line OA, and consider a line from O to the centre of the circle. What shape is made? What properties does it have?

EDIT: sorry should be the centre of the circle, not the midpoint of AB!
(edited 8 years ago)
you can make a right angled triangle with hypotenuse O to M, where M is the centre of circle C.

this hypotenuse you can express in terms of the radius r, using trig.

if you add on another r you get the total distance from O to the point on the circumference of C vertically below O.
Original post by Reety
Been stuck on part (b) of this circles question, don't quite understand it:



I get that if we split it into two, we get the angle to be pi/6, and the radius of the circle to be 6-r, but don't know what to do from there.
...
Reply 5
Original post by Charx0
I think you could use sohcahtoa (sin=o/h), and do sin(pi/6) = r/6-r
So if you do sin(pi/6) you get 1/2, so 1/2 = r/6-r
6-r = 2r and so on until r=3


I get what you're saying, but the mark scheme is saying that r=2
Original post by Reety
I get what you're saying, but the mark scheme is saying that r=2


It is 2, he made an error.
Reply 7
Original post by lizard54142
Consider a radii of the circle which is perpendicular to the line OA, and consider a line from O to the centre of the circle. What shape is made? What properties does it have?

EDIT: sorry should be the centre of the circle, not the midpoint of AB!


Original post by the bear
you can make a right angled triangle with hypotenuse O to M, where M is the centre of circle C.

this hypotenuse you can express in terms of the radius r, using trig.

if you add on another r you get the total distance from O to the point on the circumference of C vertically below O.


Original post by BuryMathsTutor
...


Aaah thanks, i understand now!
Reply 8
Original post by Reety
I get what you're saying, but the mark scheme is saying that r=2


oh sorry made a silly mistake that's what i meant, 6-r = 2r so 6 = 3r
6/3 = 2
Original post by Reety
Been stuck on part (b) of this circles question, don't quite understand it:



I get that if we split it into two, we get the angle to be pi/6, and the radius of the circle to be 6-r, but don't know what to do from there.

thanks for posting this! thought i'd try and do it, despite people already helping you, but I got stuck too :smile: was so hard not to look at what the others had put but I got there in the end :biggrin: not sure how you did it in the end, but I made a triangle like BuryMathsTutor posted then used the sine rule to find r :smile: good luck for wednesday :smile:
Reply 10
Original post by loosiemorgan
thanks for posting this! thought i'd try and do it, despite people already helping you, but I got stuck too :smile: was so hard not to look at what the others had put but I got there in the end :biggrin: not sure how you did it in the end, but I made a triangle like BuryMathsTutor posted then used the sine rule to find r :smile: good luck for wednesday :smile:


No worries, yeah i worked it out the same way as you. All the best for Wednesday!
Here is my working. Does anyone have any tips on how to spot things like this, I feel that it is common for circle / geometry questions to be like this where what you have to do is not clear at first.

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