1955 AMC 12 Problem 29

Draw $\overline{PR}$, which is tangent to both semicircles at $R$

Use the tangent-tangent angle theorem on each circle, then use $a+c=b+d=180^\circ$

The line $PR$ is tangent to both semicircles at their common endpoint $R.$ For the larger circle, the tangent-tangent angle theorem gives $$\angle APR=180^\circ-a=c.$$ For the smaller circle, the same theorem gives $$\angle RPB=180^\circ-b=d.$$ Thus the reflex angle from $\overline{PA}$ to $\overline{PB}$ through $\overline{PR}$ has measure $c+d.$ Hence the other angle between the tangents is $$\begin{aligned} 360^\circ-(c+d) &=(180^\circ-c)\\ &\quad{}+(180^\circ-d)\\ &=a+b, \end{aligned}$$ because each upper semicircle has measure $180^\circ.$

Thus, the correct answer is E.