1. ## Circle Question

I scanned the image with my visual work and I don't know where to go from there.

Any help would be greatly appreciated!

2. The $\displaystyle arc(PA)=50^o$. The angle you want intercepts $\displaystyle arc(PA)~\&~arc(BC)$.

3. How do you know that arc(DA) is 50°?

4. Originally Posted by s3a
How do you know that arc(DA) is 50°?
What is the measure of $\displaystyle arc(CBP)?$

5. Again I think you confused the letter D for P. And I don't know how to find arc(CBD). Could you tell me please?

6. Originally Posted by s3a
Again I think you confused the letter D for P. And I don't know how to find arc(CBD).
Well that could be. Your diagram is certainly shoddy.
Surely you know that a diameter creates a half-circle?

7. Hello, s3a!

$\displaystyle \Delta DAC$ is inscribed in a semicircle.
. . Hence: .$\displaystyle \angle DAC = 90^o \quad\Rightarrow\quad \angle DAB = 40^o \quad\Rightarrow\quad \text{arc}(DB) = 80^o \quad\Rightarrow\quad \angle DCB = 40^o$
In $\displaystyle \Delta BEC\!:\;\;\angle BEC + \angle ECB + \angle CBE \:=\:180^o\quad\Rightarrow\quad \angle BEC \;+\; 40^o \;+\; 65^o \;=\;180^o$
Therefore: .$\displaystyle \angle BEC \:=\:75^o$