Playing with circles and triangles

Fig

According to figure Let inradius of triangle OAB is 'r'

$r=\frac { \Delta }{ s } \\ \\ r\quad =\quad \frac { \frac { OC\times AB }{ 2 } }{ \frac { AB+OA+OB }{ 2 } } \\ \\ r\quad =\quad \quad \frac { 2\cos { \theta } \sin { \theta } }{ 2+2\cos { \theta } } \quad \quad \longrightarrow \quad (1)$ .

Now For Maximum "r"

$\frac { dr }{ d\theta } =0$ .

$\Rightarrow \quad \cos ^{ 3 }{ \theta \quad } +\quad 2\cos ^{ 2 }{ \theta } \quad -1\quad =\quad 0\\ \\ \\ \cos { \theta } =\quad -1\quad \quad \times \quad (rejected)\\ \\ or\quad \\ \\ \quad \boxed { \cos { \theta } \quad =\quad \frac { \sqrt { 5 } -1 }{ 2 } } \\$ .

Now by putting value of $\cos { \theta } \quad \& \quad \sin { \theta }$ . in expression of 'r' we get

$\Rightarrow \quad r\quad =\quad .3002$ .

Q.E.D

Let $2\theta = \angle AOB$ .

Suppose the incircle touches $OA$ at $P$ .

Then $AP=sin(\theta)$ , so $r=tan(\theta)OP = tan(\theta)(1-sin(\theta))$ .

Then just differentiate and solve to get the maximum.