+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!
+118667 Draw it up and let the side length of the square be 2. That is an easy number to work with.
Find lengths AC and AP and AQ using Pythagoras's theorem and basic logic.
What is angle ACB ?
Now it is easier to find angle CAP and then double it than it is to find angle QAP.
You can find angle CAP using either sine rule or cosine rule. (the answer will be approx.)
Edit (added): you want sine of it, not the angle itself, so that is easier and it will be exact.
Now Age26 you should have the info you need, if you want to clarify something then ask.
Please no one answer over me.
+1641 Really?!? What's your answer?
The answer to this question is: sin∠PAQ = 0.6 or 3/5
btw, your above post is ridiculous
Why do you need all that to solve such a simple question?!!
+118667 The final answer is 3/5 and I apologize for contradicting you.
However:
You did not give a solution.
A solution requires working or reasoning and you have not given that.
You pretend that this question is trivial, which it is not.
My method is correct it is just probably not the easiest way to do it.
Hectictar has given a nice solution here: It starts much as mine does.
https://web2.0calc.com/questions/geometry-help-thank-you_1#r5
Here is a quicker solution again
Again I will let the side length be 2 units because most people would find that easier than calling it by a pronumeral and it makes no difference.
Using the cosine rule
\(2=5+5-2*\sqrt5\sqrt5cos\theta\\ 2=10-10cos\theta\\ 8=10cos\theta\\ cos\theta = \frac{4}{5}\\ \text{so by drawing a simple right angled triangle with this fact is is easy to determine that} \\ sin\theta=\frac{3}{5}\)
Now jugoslav would you like to enlighten us with your trivial solution?
+118667 I see you have answered on the other post.
You said you had a trivial solution.
So why did you need to use Hectictar's picture.
Your answer is no more simple than mine or Hectictars.
You have just left out chunks to make it appear shorter.
https://web2.0calc.com/questions/geometry-help-thank-you_1#r4
+1641 I'll give you my 3-step solution without using anyone's diagram.
1/ ∠DAQ ≅ ∠BAP = arctan(DQ / AD) = 26.56505118º
2/ ∠PAQ = 90º - (∠DAQ + ∠BAP) = 36.86989765º
3/ sin(∠PAQ) = 0.6 or 3/5 
It's trivial enough, is it not?
(calculator: 0.5 => inv => tan => * 2 => -90 => + => sin => 0.6 )
From a mathematics point of view, I don't like jugoslav's answer.
It's what I would regard as an ' engineer's answer ', " close enough for all practical purposes ".
Close enough, but you can't be sure that it's exact.
It's just correct to within the accuracy limits of the calculator.
The problem arises with the use of a calculator to calculate the angle DAQ.
You might have this to however many decimal places, but evenso, all that you have is an approximation to the angle.
Follow this through, and you have to say that the final result need not be exactly 0.6 .
Melody's answer using the cosine rule is what I would regard as being mathematical.
At the end of that, I could take it as being definate that the sine of the angle PAQ is 0.6 (exactly).
Here's an alternative derivation.
Making use of Melody's diagram, call the (equal) angles DAQ and PAB, phi.
Then
\(\displaystyle \theta + 2\phi = 90^{\circ}, \\ \sin \theta = \sin(90^{\circ}-2\phi)\\ =\cos2\phi=2\cos^{2}\phi-1=2(4/5)-1= 3/5.\\ \)
