+16 In square ABCD, P is the midpoint of \(\overline{BC}\) , and Q is the midpoint of \(\overline{CD}\). Find \(\sin \angle PAQ\)
Thank you!
+9481 
Let m∠DAQ = a
Let m∠PAQ = b
Let m∠PAB = c
Notice that △DAQ ≅ △BAP, and so m∠DAQ = m∠PAB which means a = c
Then since the measure of an interior angle of a square is 90°...
a + b + c = 90°
Since a = c we can substitute a in for c
a + b + a = 90°
Combine like terms
b + 2a = 90°
Subtract 2a from both sides to solve for b
b = 90° - 2a
And so....
sin(b) = sin(90° - 2a)
Now we can use the rule that says: sin(90° - θ) = cos(θ)
sin(b) = cos(2a)
Now we can use the double-angle formula for cos
sin(b) = 1 - 2sin2(a)
And sin(a) = opposite / hypotenuse = 1 / sqrt(5) **
sin(b) = 1 - 2(1/5)
sin(b) = 3/5
**If you would like more explanation on this part please feel free to ask!
+16 Hello there hectictar, thank you so much for the reply, I've been stuck on this 1 for a while and I cant show enough support, I just have 1 question. how do we know that sin B is equal to sin(90 degrees+2a)? Thank you so much
+9481 Do you see how we know that
b = 90° - 2a ?
(If you're confused on that part, let us know!!)
Then once we know that is true, there are a couple ways to think of the next step...
Given that:
b = 90° - 2a
We can take the sin of both sides of that equation to get:
sin(b) = sin(90° - 2a)
Also......a kind of another way to think of this....
Looking at the expression:
sin(b)
Since b = 90° - 2a, we can substitute 90° - 2a in for b to get the following equivalent expression:
sin(90° - 2a)
And so....
sin(b) = sin(90° - 2a)
