In square ABCD, P is the midpoint of \(\overline{BC}\) , and Q is the midpoint of \(\overline{CD}\). Find \(\sin \angle PAQ\)

Thank you!

Age26 Oct 12, 2020

#1

#2**+4 **

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 - 2sin^{2}(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!

hectictar Oct 12, 2020

#3**0 **

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

Age26
Oct 12, 2020

#5**+3 **

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)

hectictar
Oct 13, 2020