+0

# GEOMETRY HELP!! thank you

0
46
7

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

Thank you!

Oct 12, 2020

#1
+2

The sine of angle PAQ is 0.6

Oct 12, 2020
#4
+2

By using the real numbers, we can simplify this simple question (and answer) even more. Let a side of the square be 2 units

(I'll use hecticar's diagram)

1/    ∠a ≅ ∠c = arctan(DQ / AD) ≈ 26.565º

2/    ∠b = 90º - (∠a + ∠c) ≈ 36.87º

3/     sin(∠b) = 0.6    or   3/5

jugoslav  Oct 12, 2020
edited by jugoslav  Oct 13, 2020
#7
-1

Why have you edited this?

Why don't you just be honest and be clear about your edits.

OR better still, make a new post if you have something quite different to say.

Melody  Oct 14, 2020
edited by Melody  Oct 14, 2020
#2
+3 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!

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
+2

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