Question 1186062
<pre>
{{{
drawing( 300, 300, -4, 4, -4, 4, 
         
      
         arc( 0, 0, 4, 4, 270, 0 ),
         line(0,-3, 2, 0),
         line(0,0, 2,0),
         locate(1.2,-1.5,"a"),
         locate(0.2,-1.6, alpha), 
         locate(-0.4,-1.5, "h"),
         locate(-0.5,1.2, "r"),
         locate(1.1,0.5, "r"),
         line(0,-3, 0, 2)
)
}}}

Refer to the figure above, where we've sliced the original picture down the middle.  Here {{{alpha = theta/2 }}}.   If we maximize this figure in terms of {{{alpha}}}, we just need to double that to find {{{theta}}}.

{{{A[arc]}}} = {{{ (1/4)*pi*r^2 }}}
{{{A[triangle]}}} = {{{ (1/2)r*h }}}

Observe:
{{{ r = a*sin(alpha) }}} 
{{{ h = a*cos(alpha) }}}

Substituting for r and h:
{{{A[arc]}}} = {{{ (1/4)*pi*a*sin^2(alpha) }}}
{{{A[triangle]}}} = {{{ (1/2)(a*sin(alpha))*(a*cos(alpha)) }}}

{{{ A[total] }}} = {{{A[arc] + A[triangle] }}}
= {{{ (1/4)pi*a^2*sin^2(alpha) }}} + {{{ (1/2)(a*sin(alpha))*(a*cos(alpha)) }}}

...factoring out {{{ (1/4)a^2*sin(alpha) }}}
= {{{ (1/4)(a^2*sin(alpha))* ((pi*cos(alpha) - 2sin(alpha)) ) }}}   (*)

Now take the derivative of {{{A[total]}}} with respect to {{{ alpha }}}:

{{{ dA[total] /d[alpha] }}} = 
   {{{ (1/4)a^2* (2*pi*sin(alpha)cos(alpha) + 2cos^2(alpha)-2sin^2(alpha)) }}}
= {{{ (1/4)a^2*(pi*sin(2*alpha)+2cos(2*alpha)) }}}

... substitute {{{theta = 2*alpha }}} ...
= {{{ (1/4)a^2(pi*sin(theta)+2cos(theta)) }}}  


This last function is zero at approx.  {{{theta }}} = 2.575rad, {{{theta}}}= 5.716rad, ...

By graphing, one can see 2.575rad  results in a maximum area, while 5.716 results in a minimal area.   

2.575rad is approx 147.54degrees.

Ans: {{{theta}}} = {{{147.54^o}}} maximizes the total area.

Just in case...
If you were interested in the value of the total maximal area, note that you would need a modified form of (*) because that equation computes 1/2 of the total area:
 {{{ A[total] }}} = {{{ (1/2)(a^2*sin(theta/2))* ((pi*cos(theta/2) - 2sin(theta/2)) ) }}}