Answer
(a) See explanations below.
(b) $1.05rad$ or $60.2^{\circ}$
Work Step by Step
(a) Use the figure given with the Exercise, the length of the arc on the left wheel is $L1=R\Phi$ where $Phi=2\pi-(\pi-\theta)=\pi+\theta$. The length of the straight parts on the left side of the cross point is $L2=2Rcot\frac{\theta}{2}$. Similarly for the right wheel, $L3=r(\pi+\theta)$ and $L4=2r\cdot cot\frac{\theta}{2}$.
Since the total length is $L$, we have $L=(R+r)(\pi+\theta)+2(R+r)cot\frac{\theta}{2}$. Divide both sides of the equation by $R+r$, we get $\frac{L}{R+r}=\pi+\theta+2cot\frac{\theta}{2}$ which is the same as the equation given in the Exercise.
(b) Plug-in the number in the above equation, we have $\theta+2cot\frac{\theta}{2}=\frac{27.78}{2.42+1.21}-\pi\approx4.5113$ Graph this function as shown, we can find a zero at $\theta\approx1.05rad\approx60.2^{\circ}$
