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Mathematics 27 Online
danielaaxrosas89:

Find the value of x and y. Write your answer in simplest form.

Shadow:

Is there an equation or image that relates to x and y?

danielaaxrosas89:

Shadow:

@iGreen You want this one? :)

Shadow:

hi @danielaaxrosas89 Have you heard of the Law of Sines?

danielaaxrosas89:

I don’t think so?

iGreen:

@danielaaxrosas89 I'm going to let @Shadow help you out on this one, I've got HW to do.

danielaaxrosas89:

It’s okay.

danielaaxrosas89:

@iGreen

Shadow:

It basically goes like this: \[\frac{ \sin A}{ a }= \frac{ \sin B }{ b } = \frac{ \sin C }{ c }\] Where C is angle C, and c is side c

Shadow:

You basically use this to compare sides and angles in order to solve for unknowns.

Shadow:

|dw:1519443720911:dw|

danielaaxrosas89:

Would it be 45 to the other side?

Shadow:

The other angle?

danielaaxrosas89:

Yes

Shadow:

Yes, you are correct

Shadow:

That information is vital in order to solve this problem, as in order to use Law of Sines, you need three pieces of information. The value of an angle and it's corresponding side, as well as any side or angle. So now we can do this \[\frac{ \sin 45 }{ 5 } = \frac{ \sin 45 }{ y }\]

Shadow:

Do you see how I got that setup?

danielaaxrosas89:

So “y“ would be 5?

Shadow:

Yes

Shadow:

And that makes sense, since in 45, 45, 90 triangles, the sides opposite to the 45 degree angles are of the same length.

Shadow:

Now since we have two sides, we can do two things. We can continue with Law of Sines (and get some practice in) or do Pythagoras' Theorem. What do you want to try?

danielaaxrosas89:

Probably Pythagorean theorem?

Shadow:

Either one would work

danielaaxrosas89:

Alright, let’s do Pythagorean theormen

Shadow:

Set it up :)

Shadow:

Do you know how? Or do you need help?

danielaaxrosas89:

I think I need help, because I don’t know which #’s to plug in, since there are 4 of them

Shadow:

Well Pythagorean Theorem only deals with sides. So far we have solved for two of them. Both of the sides have a value of 5. @iGreen Had a good description of Pythagorean's Theorem in the previous post in which he worked with you. \[a^2 + b^2 = c^2\] 'a' and 'b' are the legs of the triangle. In this case, they would both have a value of five. c is the hypotenuse, which is the side always opposite to the right angle, or 90 degree angle.

Shadow:

|dw:1519444816429:dw|

Shadow:

We know that y = 5

Shadow:

Do you think you could set it up now?

danielaaxrosas89:

I got 7.07 ?

Shadow:

\[a^2 + b^2 = c^2\] \[5^2 + 5^2 = x^2\] \[25 + 25 = x^2 \rightarrow 50 = x^2\] \[\sqrt 50 = \sqrt x^2 \rightarrow \sqrt 50 = x\] \[\sqrt (25 \times 2) = x \] \[5 \sqrt 2 = x\] If you multiply 5 by sqrt 2, you get 7.07

Shadow:

So you are correct

danielaaxrosas89:

So will my answer be x=7.07 , y=5 or will both my x and y be 7.07 ?

Shadow:

Not sure if you learned this already, but there is something called Special Right Triangles These are the 30 - 60 - 90 and 45 - 45 - 90 triangles. It basically means the following: |dw:1519445241235:dw| If you can solve for x, you can get the hypotenuse immediately.

Shadow:

This is when two angles are 45

Shadow:

Remember when we used the Law of Sines. y was the length opposite to a 45 degree angle. We used Law of Sines and got 5. Also, if you use Special Right Triangles, you also get 5 for y.

Shadow:

x was the hypotenuse, and we used Pythagorean Theorem to solve for it. We got \[5 \sqrt 2 \] which is equal to 7.07, so either one works.

Shadow:

Though most teachers prefer \[5 \sqrt 2\] but that's just in my experience. If your teacher asks for a decimal, then give 7.07

danielaaxrosas89:

I have learned special right triangles, in fact. So I can put either a or b as my answer? If you look at the given answers picture.

Shadow:

Put this: \[y = 5, x = 5 \sqrt 2\]

danielaaxrosas89:

a or d*

1 attachment
Shadow:

I think we are on d

danielaaxrosas89:

Alright

danielaaxrosas89:

Well thanks for your help! I think I understand how to solve for x and y. I might have to ask you guys for help again, later on! 😅

Shadow:

Okay, I'll be on : )

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