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Maths Problem Please Help

In the following diagram, O is the centre of the circle, OA and BC are parallel, D is the intersection of AB and OC. If angle OCB is 40 degrees, find angle ODA.

I am unsure of this question and it is worth six marks. I would be very grateful for any help, however if you do give an answer please could you show your working.
Thank you

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IMG_5340.jpg
Have you tried solving it?
Yes
A quick glance but I think it's 120. Is that the answer? If so I can explain with hints.
Could you elaborate as why you think it is 120 degrees?
I have tried he question a few times but I know the answers I have gotten are wrong.
Any explanation would be appreciated 😄
Original post by Charleslewison
Could you elaborate as why you think it is 120 degrees?
I have tried he question a few times but I know the answers I have gotten are wrong.
Any explanation would be appreciated 😄


its 100 , he's wrong
Original post by Ollie123x
its 100 , he's wrong


Thankyou, please could you explain why it is 100😄
join O to B & look at the triangle OBC.

also look at the triangle OBA
Original post by the bear
join O to B & look at the triangle OBC.

also look at the triangle OBA


So in triangle OBC would angle BOC be 90 degrees because of Theorem 2?
Original post by Charleslewison
So in triangle OBC would angle BOC be 90 degrees because of Theorem 2?


BC is not a diameter and O is not on the edge of the circle, so no.

the triangle is isosceles....
Original post by the bear
BC is not a diameter and O is not on the edge of the circle, so no.

the triangle is isosceles....


Oh I see that now😅 so angle OCB is 40 degrees as is angle OBC and therefore angle BOC would be 100 degrees?
Original post by Charleslewison
IMG_5340.jpg


Do you just want angle D, on the bottom triangle?
Original post by Charleslewison
Oh I see that now😅 so angle OCB is 40 degrees as is angle OBC and therefore angle BOC would be 100 degrees?


yes !
Original post by Ollie123x
Do you just want angle D, on the bottom triangle?


No I think the question is asking for angle ODA on the top triangle, the smaller one.
(Sorry I have just realised the photograph is upside down😅)
Original post by Charleslewison
Oh I see that now😅 so angle OCB is 40 degrees as is angle OBC and therefore angle BOC would be 100 degrees?


Alternate angle theorem makes COA , 40 degrees that is certain as BC and OA are parallel. I may be mistaken but the bow tie circle theorem allows you to deduce that angle BAO is also 40 which makes the final angle which you want 100...?
Original post by Charleslewison
No I think the question is asking for angle ODA on the top triangle, the smaller one.
(Sorry I have just realised the photograph is upside down😅)

They're the same anyway :smile:
Not only do you give us a difficult math question (For me), but you give it to us upside down? Gheez gheez, settle down Einstein. Is this what makes a grade 9 question?
Original post by Ollie123x
Alternate angle theorem makes COA , 40 degrees that is certain as BC and OA are parallel. I may be mistaken but the bow tie circle theorem allows you to deduce that angle BAO is also 40 which makes the final angle which you want 100...?


Thank you for the help! 😄 Just out of curiosity would angle BDC be 100 as degrees as well?
Original post by Charleslewison
Thank you for the help! 😄 Just out of curiosity would angle BDC be 100 as degrees as well?

I believe so, message me if you need any more help

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