A surprising answer

Let radius of circle with center $B$ be $R$ and circle with center $O$ be $r$ .

In $\triangle OBC$ ,

$OB^2=OC^2+BC^2$ ......... Pythagoras' theorem

$\therefore \underbrace{(R+r)^2}_{\text{Property of touching circles}} = \underbrace{(R-r)^2}_\text{OC=CD-OD=R-r} + R^2\\ \therefore R^2+2Rr+r^2= R^2-2Rr+r^2 +R^2\\ \therefore R^2=4Rr \\ \therefore \dfrac{r}{R}=\dfrac{1}{4} \\ \implies \dfrac{R+r}{R}=\dfrac{1+4}{4} \quad \quad\text{Adding 1 on both sides} \\ \therefore \dfrac{OB}{BC}=\dfrac{5}{4} \\ \therefore \cos \angle OBC = \dfrac{4}{5}$

Hence answer is $4+5=\boxed{9}$

I liked Adiya's way, but my way is informal, lets take the radius of big circle be 1 and that of small circle be R . In right triangle OBC, $OB=1+R, OC=1-R$ Using Pythagoras, $R=\frac{1}{4}$ Now $\cos\angle OBC=\frac{1}{1+1/4}=\frac{4}{5}$ Ans $4+5=\boxed9$

Hi! Remember that you only need 4 lines of DO to equal 1 line BC

Just the pythagorean! BC = 4, OC = 3 , and BO = 5! now just put the requirements 4+5 = 9 :P

What is surprising?

Let the large and small circle be respectively:

$Circle(B, R);\;Circle(O, r);$

Let $\;\theta = \angle OBC,\;$ and let $T$ be the tangent point between circles; we have:

$|BO| = |BT| + |TO| = R + r;$

$|BC| = R;$

$|CO| = |CD| - |OD| = R - r;$

by Pytagoras’ theorem:

$|BC|^{ 2 } + |CO|^{ 2 } = |BO|^{ 2 };$

$R^{2} + (R-r)^{2} = (R+r)^{2};$

simplyfing:

$R(R-4r) = 0;\; R = 4r;$

$cos\theta = \frac{|BC|}{|BO|} = \frac{R}{R + r} = \frac{4r}{4r + r} = \frac{4r}{5r} = \frac 4 5 = \frac a b;$

$a + b = 4 + 5 = \boxed 9;$

$\theta = arccos(\frac{4}{5}) = 36.86989765\;deg$