# Mathswatch proof of circle theorems

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#1
question 4
please anyone know how to do this??
0
1 year ago
#2
(Original post by Afif8011)
question 4
please anyone know how to do this??
Consider the angle OBQ - what is is?

Hence what's OBC?

And carry on.
1
#3
Sorry this time calculated both but no marks were given, do you have any full answer, would really appreciate that.
(Original post by ghostwalker)
Consider the angle OBQ - what is is?

Hence what's OBC?

And carry on.
0
1 year ago
#4
(Original post by Afif8011)
Sorry this time calculated both but no marks were given, do you have any full answer, would really appreciate that.
No, I don't have a full answer; and it's against forum rules.

You can post working if you'd like it checked
1
#5
(Original post by ghostwalker)
No, I don't have a full answer; and it's against forum rules.

You can post working if you'd like it checked
Yes please, how do I improve this?
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1 year ago
#6
(Original post by Afif8011)
Yes please, how do I improve this?
Well OBC doesn't equal x. You're perhaps being mislead by the inclination of OC; remember the diagram isn't drawn accurately. See my first post on determining CBO.
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#7
(Original post by ghostwalker)
Well OBC doesn't equal x. You're perhaps being mislead by the inclination of OC; remember the diagram isn't drawn accurately. See my first post on determining CBO.
Ahh well I'm struggling to get my head around this, I don't know what OBC or the rest are equal to
0
1 year ago
#8
(Original post by Afif8011)
Ahh well I'm struggling to get my head around this, I don't know what OBC or the rest are equal to
OBQ = 90 = OBC + CBQ

And you know what CBQ is from the diagram. Hence....
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#9
(Original post by ghostwalker)
OBQ = 90 = OBC + CBQ

And you know what CBQ is from the diagram. Hence....
tried, doesn't work bro
0
1 year ago
#10
(Original post by Afif8011)
tried, doesn't work bro
Sorry, should have said CBO instead of OBC, although it's the same angle.

The question asks you to work out 4 angles in terms of x, before showing that y=2x

You have DCO correctly.

Then.

OBQ = 90 = CBO + CBQ

So, what's CBO in terms of x (it will also include some number(s)), given that CBQ is 2x?
0
#11
(Original post by ghostwalker)
Sorry, should have said CBO instead of OBC, although it's the same angle.

The question asks you to work out 4 angles in terms of x, before showing that y=2x

You have DCO correctly.

Then.

OBQ = 90 = CBO + CBQ

So, what's CBO in terms of x (it will also include some number(s)), given that CBQ is 2x?
Still doesn't work :/, but thanks for trying
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1 year ago
#12
(Original post by Afif8011)
Still doesn't work :/, but thanks for trying
Why do you think CBO is equal to x?
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#13
I'm just dumb lol, idk what it is
(Original post by ghostwalker)
Why do you think CBO is equal to x?
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1 year ago
#14
It's not quite CBO that's x, look at the radii and think about isoceles triangles. Which one is the same base as angle x?
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#15
I think DCO and OCB equals to x as isosceles triangle has two equal angles
(Original post by GiveMeCoffee4)
It's not quite CBO that's x, look at the radii and think about isoceles triangles. Which one is the same base as angle x?
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1 year ago
#16
DCO is definitely x. Before you find OCB, you need to find OBC. Think about angles from radii to tangents
(Original post by Afif8011)
I think DCO and OCB equals to x as isosceles triangle has two equal angles
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1 year ago
#17
(Original post by Afif8011)
I think DCO and OCB equals to x as isosceles triangle has two equal angles
The two isosceles triangles are not congurent. They have different angles.

Your reasoning for angle DCO is correct. You're given the angle ODC and this equals the angle DCO.

You can't use any of that information for the triangle OCB, as it's a completely different triangle with different angles.

What you do know is OBQ is a right angle, i.e. equals 90 degrees, and CBQ = 2x.

Hence looking at the diagram angle OBC = 90 - angle CBQ = ...
0
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