$\angle AOB=75°$, $\angle CBD=62°$, $\angle BAD=30°$ find $\angle BDA$ and $\angle ABD$

Question

I was looking at some geometry examples and struggling to solve this one:

Given that $O$ is the center of the circle and that $\angle AOB=75°$, $\angle CBD=62°$, and $\angle BAD=30°$ calculate:

a) $\angle BDA$

b) $\angle ABD$

So far I have found $\angle ACB=37.5°$

Can you please help?

Answer 1

Note the

a) $\angle BDA=\frac{1}{2}\overparen{AB}=\frac{1}{2}\angle AOB=37,5º$

and

b) $\angle ABD=\frac12\overparen{ACD}=\frac12(360º-\overparen{AB}-\overparen{BD})=\frac12(360º-\angle AOB-2.\angle BAD)=112,5º$.

For details of this properties see https://mathbitsnotebook.com/Geometry/Circles/CRAngles.html

Answer 2

For the second part note that $$180^\circ=\angle ABD+\angle BDA+\angle BAD$$ and you have already found $\angle BDA$, and it is given that $\angle BAD=30^\circ$.