Angle relations Please can anyone help finding the solution.. I am working on this task in geometry and already found out some angles being $90^{\circ}$ and $17^{\circ}$ and $34^{\circ}$. I tried through the sum of the angles:
$$360^{\circ} = 197^{\circ} - \alpha + 90^{\circ} + 73^{\circ} + \alpha$$
but always didn't work out. can anyone please help me?

 A: Construct the line joining centre of bigger circle to the point of intersection of circumference of small circle and bigger circle. Use the fact that angle at centre is double of angle and circumference, base angles of isosceles triangle are equal and the angle subtended by same arc on circumference is equal to get answer 51(??). I will post a picture if possible.

A: Denote the top point $P$, the bottom $Q$ and those on the horizontal diameter $A, B, C, D$ from left to right.
We are given that $\angle PDC=\angle PDA=17^\circ.$
Then 
$$\begin{align}\angle PQA&=\angle PDA=17^\circ&\text{(peripheral angle),}\\
\angle PCB&=\angle PCA=2\angle PDA=34^\circ&\text{(central angle),}\\
\angle BPC&=\angle PCB=34^\circ&\text{(isosceles triangle),}\\
\angle CPD&=\angle BPD-\angle BPC=\alpha-34^\circ&\text{(difference),}\\
\angle DCP&=180^\circ -\angle PDC-\angle CPD=197^\circ-\alpha&\text{(sum in triangle).}\end{align}$$
From the straight angle $\angle DCA$ we find
$$180^\circ = \angle DCP+\angle PCA=34^\circ +197^\circ-\alpha,$$
i.e.  $$\alpha=51^\circ.$$
