Question 1207589
<font color=black size=3>
This is possibly what the diagram looks like
{{{
drawing(400,400,-3,3,-3,3,

locate(2,-1,"2"),locate(2,0.5,"3"),
locate(-0.7,2,"3"),locate(1.1,2,"2"),
locate(0,-1.7,"5"),locate(-1.7,0,"5"),

locate(-1.3-0.3,-1.46-0.2,"A"),locate(1.7,-1.46-0.2,"B"),locate(1.7+0.1,1.54,"C"),locate(-1.3-0.3,1.54+0.2,"D"),locate(1.7+0.1,-0.46,"P"),locate(0.7-0.2,1.54+0.3,"Q"),

circle(-1.4,-1.56,0.05),circle(-1.4,-1.56,0.07),circle(-1.4,-1.56,0.09),circle(-1.4,-1.56,0.11),circle(1.6,-1.56,0.05),circle(1.6,-1.56,0.07),circle(1.6,-1.56,0.09),circle(1.6,-1.56,0.11),circle(1.6,1.44,0.05),circle(1.6,1.44,0.07),circle(1.6,1.44,0.09),circle(1.6,1.44,0.11),circle(-1.4,1.44,0.05),circle(-1.4,1.44,0.07),circle(-1.4,1.44,0.09),circle(-1.4,1.44,0.11),circle(1.6,-0.56,0.05),circle(1.6,-0.56,0.07),circle(1.6,-0.56,0.09),circle(1.6,-0.56,0.11),circle(0.6,1.44,0.05),circle(0.6,1.44,0.07),circle(0.6,1.44,0.09),circle(0.6,1.44,0.11),

line(-1.4,-1.56,1.6,-1.56),line(1.6,-1.56,1.6,1.44),line(1.6,1.44,-1.4,1.44),line(-1.4,1.44,-1.4,-1.56),line(-1.4,-1.56,0.6,1.44),line(-1.4,-1.56,1.6,-0.56),

locate(-2.5,-2.5,matrix(1,4,"Diagram","not","to","scale"))

)
}}}


Focus on right triangle PAB.
We'll use the tangent trig ratio to determine the following.
tan(angle) = opposite/adjacent
tan(angle PAB) = PB/AB
tan(angle PAB) = 2/5
angle PAB = arctan(2/5)
angle PAB = 21.801409 degrees approximately.


Now move your attention to right triangle DAQ.
tan(angle) = opposite/adjacent
tan(angle DAQ) = DQ/AD
tan(angle DAQ) = 3/5
angle DAQ = arctan(3/5)
angle DAQ = 30.963757 degrees approximately.


Then,
(angleDAQ) + (anglePAQ) + (anglePAB) = angleDAB
(angleDAQ) + (anglePAQ) + (anglePAB) = 90
anglePAQ = 90 - (angleDAQ + anglePAB)
anglePAQ = 90 - (30.963757 + 21.801409)
anglePAQ = 37.234834 degrees approximately
sin(angle PAQ) = sin(37.234834) = <font color=red>0.605083 approximately</font>



------------------------------------------------------------------------------------


Another approach.


Use the Pythagorean Theorem to determine these lengths
QA = sqrt(5^2+3^2) = 5.830952 approximately
PA = sqrt(5^2+2^2) = 5.385165 approximately
QP = sqrt(2^2+3^2) = 3.605551 approximately


Focus on triangle PAQ of the diagram shown above.
Use the Law of Cosines to find angle PAQ.
a^2 = b^2+c^2-2*b*c*cos(A)
(QP)^2 = (QA)^2+(PA)^2-2*(QA)*(PA)*cos(angle PAQ)
(3.605551)^2 = (5.830952)^2+(5.385165)^2-2*(5.830952)*(5.385165)*cos(angle PAQ)
12.999998 = 63.000003 -62.801277*cos(angle PAQ) 
12.999998 - 63.000003 = -62.801277cos(angle PAQ)
-50.000005 = -62.801277cos(angle PAQ)
cos(angle PAQ) = -50.000005/(-62.801277)
cos(angle PAQ) = 0.796162
angle PAQ = 37.234852
There appears to be some slight rounding error going on. 
Compare this value to the previous angle PAQ measure found in the section above.
</font>