Question 1207455
I just ignored "Sin".
<pre>  
This problem would be much easier if you could use decimals, but 
decimals are approximate. The answer could only be approximate if we 
use decimals, so we must use exact values.  Maybe some other tutor can 
come up with a shorter solution, but this was the very long solution I 
came up with:

{{{drawing(400,1200/11,-.1,2.1,-.1,.5,

locate(.3,.13,15^o),locate(1.7,.13,30^o),
locate(0,0,A),locate(2,0,B),
locate(1.35,.45,C), locate(1,0,M),
locate(.5,0,1), locate(1.5,0,1),
locate(1.15,.3,theta),

green(line(1.366025404,.366025404,1,0)),
locate(.6,.25,b), locate(1.7,.25,a),
triangle(0,0,2,0,1.366025404,.366025404) )}}}

We want to find angle &theta; 

Without loss of generality we can let AM = MB = 1 unit.

We are going to need expressions for both sine and cosine of 15<sup>o</sup>.

{{{sin(15^o)}}}{{{""=""}}}{{{sin(45^o-30^o)}}}{{{""=""}}}{{{sin(45^o)cos(30^o)-cos(45^o)sin(30^o)}}}{{{""=""}}}{{{(sqrt(2)/2)(sqrt(3)/2)-(sqrt(2)/2)(1/2)}}}{{{""=""}}}{{{sqrt(6)/4-sqrt(2)/4}}}{{{""=""}}}{{{(sqrt(6)-sqrt(2))/4}}}

{{{cos(15^o)}}}{{{""=""}}}{{{cos(45^o-30^o)}}}{{{""=""}}}{{{cos(45^o)cos(30^o)+sin(45^o)sin(45^o)}}}{{{""=""}}}{{{(sqrt(2)/2)(sqrt(3)/2)+(sqrt(2)/2)(1/2)}}}{{{""=""}}}{{{sqrt(6)/4+sqrt(2)/4}}}{{{""=""}}}{{{(sqrt(6)+sqrt(2))/4}}}

Angle ACB = 180<sup>o</sup>-15<sup>o</sup>-30<sup>o</sup> = 135<sup>o</sup>.

By the law of sines on triangle ABC.

{{{AB/sin(ACB)}}}{{{""=""}}}{{{AC/sin(B)}}}{{{""=""}}}{{{BC/sin(A)}}}

{{{AB/sin(135^o)}}}{{{""=""}}}{{{b/sin(30^o)}}}{{{""=""}}}{{{a/sin(15^o)}}}

{{{2^""/(1/sqrt(2))}}}{{{""=""}}}{{{b^""/(1/2)}}}

{{{2*sqrt(2)}}}{{{""=""}}}{{{2b}}}

{{{sqrt(2)}}}{{{""=""}}}{{{b}}}

{{{b/sin(30^o)}}}{{{""=""}}}{{{a/sin(15^o)}}}

{{{sqrt(2)^""/(1/2)}}}{{{""=""}}}{{{a^""/((sqrt(6)-sqrt(2))/4))}}}

{{{2*sqrt(2)}}}{{{""=""}}}{{{4a/(sqrt(6)-sqrt(2))}}}

{{{sqrt(2)}}}{{{""=""}}}{{{2a/(sqrt(6)-sqrt(2))}}}
{{{sqrt(2)(sqrt(6)-sqrt(2)^"")}}}{{{""=""}}}{{{2a}}}
{{{sqrt(12)-2}}}{{{""=""}}}{{{2a}}}
{{{sqrt(4*3)-2}}}{{{""=""}}}{{{2a}}}
{{{2sqrt(3)-2}}}{{{""=""}}}{{{2a}}}
{{{sqrt(3)-1}}}{{{""=""}}}{{{a}}}

By the law of cosines on triangle ACM

{{{CM^2}}}{{{""=""}}}{{{AC^2+AM^2-2*AC*AM*cos(15^o)}}}
{{{CM^2}}}{{{""=""}}}{{{b^2+1^2-2*b*1*((sqrt(6)+sqrt(2))/4)}}}
{{{CM^2}}}{{{""=""}}}{{{(sqrt(2))^2+1-2*sqrt(2)*1*((sqrt(6)+sqrt(2))/4)}}}
{{{CM^2}}}{{{""=""}}}{{{2+1-sqrt(2)((sqrt(6)+sqrt(2))/2)}}}
{{{CM^2}}}{{{""=""}}}{{{3-expr(1/2)*sqrt(2)(sqrt(6)+sqrt(2)^"")}}}
{{{CM^2}}}{{{""=""}}}{{{3-expr(1/2)*(sqrt(12)+2^"")}}}
{{{CM^2}}}{{{""=""}}}{{{3-expr(1/2)*(sqrt(4*3)+2^"")}}}
{{{CM^2}}}{{{""=""}}}{{{3-expr(1/2)*(2*sqrt(3)+2^"")}}}
{{{CM^2}}}{{{""=""}}}{{{3-sqrt(3)-1}}}
{{{CM^2}}}{{{""=""}}}{{{2-sqrt(3)}}}
{{{CM}}}{{{""=""}}}{{{sqrt(2-sqrt(3))}}}

Now let's see if we can rewrite that so it doesn't have a 
square root under a square root.  We'll see if we can write
it as the difference of 2 square roots:

{{{sqrt(2-sqrt(3))}}}{{{""=""}}}{{{sqrt(P)-sqrt(Q)}}}
{{{2-sqrt(3)}}}{{{""=""}}}{{{P-2sqrt(PQ)+Q}}}
{{{system(2=P+Q,-sqrt(3)=-2sqrt(PQ))}}}=>{{{system(Q=2-P,4PQ=3)}}}
{{{4P(2-P)}}}{{{""=""}}}{{{3}}}
{{{8P-4P^2}}}{{{""=""}}}{{{3}}}
{{{0}}}{{{""=""}}}{{{4P^2-8P+3}}}
{{{0}}}{{{""=""}}}{{{(2P-1)(2P-3)}}} => {{{matrix(1,3,P=1/2, or, P=3/2)}}} => {{{matrix(1,3,Q=3/2, or, Q=1/2)}}}
CM can't be negative, so {{{matrix(1,3,P=3/2, and, Q=1/2)}}}
{{{CM}}}{{{""=""}}}{{{sqrt(3/2)-sqrt(1/2)}}}{{{""=""}}}{{{sqrt(6/4)-sqrt(2/4)}}}{{{""=""}}}{{{ sqrt(6)/2-sqrt(2)/2}}}
{{{CM}}}{{{""=""}}}{{{(sqrt(6)-sqrt(2))/2}}}

Use the law of sines on triangle ACM:

{{{AM/sin(theta)}}}{{{""=""}}}{{{CM/sin(A)}}}
{{{1/sin(theta)}}}{{{""=""}}}{{{((sqrt(6)-sqrt(2))/2)/((sqrt(6)-sqrt(2))/4)}}}
{{{1/sin(theta)}}}{{{""=""}}}{{{4/2}}}{{{""=""}}}{{{2}}}
{{{sin(theta)}}}{{{""=""}}}{{{1/2}}}
{{{theta}}}{{{""=""}}}{{{30^o}}}

Edwin</pre>