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Circle theorems help

Can someone help me how to find angle BDC and OCD
(edited 6 years ago)
Screenshot_20171025-210245.png103.0KB
Original post by Chocolate...s
Can someone help me how to find angle BDC and OCD


Unless I'm being thick or forgotten something, I don't see how those angles can be obtained from the given information.

Where did you find this Q?
Reply 2
Avatar for B_9710
B_9710
18
Original post by Chocolate...s
Can someone help me how to find angle BDC and OCD


ACO is 90 as tangent and radius are perpendicular, from there you can easily work out AOC and then BDC follows as angle at centre is 2 x angle at circumference.
The key here is just to work out each angle and make and put it on the diagram - you should be able to work out all angles.
Original post by RDKGames
Unless I'm being thick or forgotten something, I don't see how those angles can be obtained from the given information.

Where did you find this Q?


ahem... the angle at the centre is double the angle at the edge

:console:
Original post by B_9710
ACO is 90 as tangent and radius are perpendicular, from there you can easily work out AOC and then BDC follows as angle at centre is 2 x angle at circumference.
The key here is just to work out each angle and make and put it on the diagram - you should be able to work out all angles.


I wouldn’t have gone the way you went, I saw that ABDC is a parallelogram, meaning that opposite angles are equal BCD is 32 and therefore OCD would be ((360 -(32*2))-90)
Original post by RDKGames
Unless I'm being thick or forgotten something, I don't see how those angles can be obtained from the given information.

Where did you find this Q?

https://corbettmaths.files.wordpress.com/2013/02/circle-theorems.pdf
Question 20
Original post by the bear
ahem... the angle at the centre is double the angle at the edge

:console:


Ah, I know the rule but I was struggling to apply it but after 'untangling' the quadrilateral BOCD it's a smack in the face
Original post by RDKGames
Ah, I know the rule but I was struggling to apply it but after 'untangling' the quadrilateral BOCD it's a smack in the face


https://shechive.files.wordpress.com/2014/02/k.jpg?quality=100&strip=info
Original post by the bear


Luckily I didn't have to use my trusty rotating tool there.
Original post by Daniel100499
I wouldn’t have gone the way you went, I saw that ABDC is a parallelogram, meaning that opposite angles are equal BCD is 32 and therefore OCD would be ((360 -(32*2))-90)


The diagram is deceptive - ABDC is not a parallelogram. Although one pair of sides are parallel, the other pair are not.

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