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.
ANSWER:
Best Answer
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 $y$.