Question 1207589
<pre>

Maybe you want exact values, not decimal approximations.

{{{drawing(400,400,-1,6,-1,6,

triangle(0,0,0,5,2,0),
triangle(2,0,5,0,5,2),
triangle(0,5,5,5,5,02),

locate(0,0,B), locate(5,0,C),locate(0,5.3,A), locate(5,5.3,D),

locate(1,0,2), locate(3.5,0,3), locate(5.1,1,2), locate(5.1,3.5,3) 

locate(2,0,P), locate(5.1,2,Q),
red(
locate(-.2,2.5,5), locate(2.5,5.3,5), locate(3.52,1.1,sqrt(13)),

locate(2.5,3.8,sqrt(34)),

locate(.6,2.3,sqrt(29)) ) )}}} 

Use the properties of a square and the Pythagorean theorem
on the 3 right triangles to get all the measures in red.

Then use the law of cosines to find the cos(PAQ):

{{{PQ^2}}}{{{""=""}}}{{{AP^2+AQ^2-2*AP*AQ*cos(PAQ)}}}

{{{sqrt(13)^2}}}{{{""=""}}}{{{sqrt(29)^2+sqrt(34)^2-2*sqrt(29)*sqrt(34)*cos(PAQ)}}}

{{{13}}}{{{""=""}}}{{{29+34-2*sqrt(34)*sqrt(29)*cos(PAQ)}}}

{{{13}}}{{{""=""}}}{{{63-2*sqrt(986)cos(PAQ)}}}

{{{2*sqrt(986)*cos(PAQ)}}}{{{""=""}}}{{{63-13}}}

{{{2*sqrt(986)*cos(PAQ)}}}{{{""=""}}}{{{50}}}

{{{cos(PAQ)}}}{{{""=""}}}{{{50/(2*sqrt(986))}}}

{{{cos(PAQ)}}}{{{""=""}}}{{{25/sqrt(986)}}}


Then use the identity

{{{sin^2(PAQ)}}}{{{""=""}}}{{{1-cos^2(PAQ)}}}

{{{sin^2(PAQ)}}}{{{""=""}}}{{{1-(25/sqrt(986))^2}}}

{{{sin^2(PAQ)}}}{{{""=""}}}{{{1-625/986}}}

{{{sin^2(PAQ)}}}{{{""=""}}}{{{361/986}}}

{{{sin(PAQ)}}}{{{""=""}}}{{{sqrt(361/986)}}}

{{{sin(PAQ)}}}{{{""=""}}}{{{sqrt(expr(361/986)*expr(986/986))}}}

{{{sin(PAQ)}}}{{{""=""}}}{{{sqrt(361*986)/986))}}}

{{{sin(PAQ)}}}{{{""=""}}}{{{(19*sqrt(986))/986))}}}

That is approximately 0.6050832675, which agrees with the other tutor.

Edwin</pre>