Proving algebraic equations with circle theorems

Question

I got as far as stating that OBP=90˚ (as angle between tangent and radius is always 90˚), and thus CBO=90˚- 2x. CBO=OCB as they are bases in a isosceles. COB=180-90-2x-90-2x. But after this, i am clueless.

I am stuck with this Question. It is from a GCSE Further Maths past paper. Despite seeing online tutorials, and checking the answer scheme, I still don't understand how you solve this question. Could you please show me a step by step explanation of how you solve this question. Thank you.

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ANSWER:

Answer 1

Angle BOD = 180 − y

Angle OCD = x

Angle OBC = 90 − 2x

Angle BCO= 90 − 2x

Angle BOD reflex = 360 - (90 − 2x) − (90 − 2x) − x − x = 180 + 2x.

180 − y + 180 + 2x = 360,

thus y = 2x

Answer 2

Try expressing the same angle both in terms of $x$ and in terms of $y$. Then you can set the expressions equal to each other and rearrange. For example, you can do this with either $\angle BOD$ (either the reflex or obtuse angle) or $\angle BCD$. Does this help?

Answer 3

At first forget about the condition that $\angle CBP = 2x$. (The condition $\angle CBP = 2 \angle ODC$ determines the exact position of C.)

You can express $\angle BCD$ with $y$. Using $\angle OCD = \angle ODC = x$ you can express $\angle OCB = OBC$ with $x$ and $y$ and hence also $\angle CBP$.

Now use the condition that $\angle CBP$ equals $2x$. This gives an equation in $x$ and $y$.

Answer 4

I am also doing Further Maths GCSE and what i would do is draw a line from B to D (this is a lot shorter) this would create the alternate segment theorem for angles therefore $BDC$ would be $2x$ and $BDO$ and $DBO = x$ which means $ BOD = 180- 2x$ which means $y = 360 − 90 − 90 − (180 − 2x)$

$y = 2x$