help please

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- purplemexican

help please

- schrodinger

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- purplemexican

##### 1 Attachment

- tkhunny

Please provide the Right-Triangle definition of the Cosine.

- purplemexican

everywhere

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## More answers

- purplemexican

the most i can tell you is what im looking for

- tkhunny

You cannot have been given this problem without the basic Right-Triangle definitions of Sine, Cosine, and Tangent. What are they?

- purplemexican

do i know what?

- purplemexican

i have been given what you see

- purplemexican

and 4 answer choices

- purplemexican

if i knew earlier today i dont remember now

- tkhunny

Hogwash, unless this is a placement test. If this is a placement test, you should get this problem wrong so that you are placed in the correct course this semester.

- purplemexican

well i dont want you to do it for me i would like a step by step tho

- purplemexican

im in summerschool

- purplemexican

ive been doing algebra 2 since 9 am to 4:30 for the past few days im half out of my wits with this stuff now

- tkhunny

So? Then take a break. Eat something. Get some sleep.
Read this: https://www.bing.com/images/search?q=right+triangle+definition+of+sine&view=detailv2&&id=215DE835C6C9753A48093FB88A6FD84396447F91&selectedIndex=0&ccid=6mvOJUX%2f&simid=608044099930951339&thid=JN.N04zAIXgt9vGSoV1a0kegw&ajaxhist=0

- purplemexican

i have till 4:30 tomorrow to finish about 50 questions i would love to take a brake i really wold

- tkhunny

Did you read it?

- purplemexican

yes it make sence

- tkhunny

Okay, now solve. What does it want and how shall you find it?

- purplemexican

i have one minuet im gonna guess on the quiz then come back and figure it out

- purplemexican

didnt even get to guess so im looking for angle FHE right and or cos(theta) right?

- tkhunny

\(\cos(\angle FHE) = \dfrac{Adjacent}{Hypotenuse}\)
You know neither the Adjacent Side (FH) nor the Hypotenuse (EH). How shall we find them?

- purplemexican

i have 2 angles and a side i could find the other sides of the one triangle and then i would have one of the sides

- tkhunny

Good. How about \(\triangle{HGF}\)? Can that help us find FH?

- purplemexican

yes it would

- tkhunny

It's a 45-45-90 Right Triangle. What is the relationship between the Leg and the Hypotenuse?

- purplemexican

leg FH is 4 right

- tkhunny

How did you get that?
One for free: \(\sqrt{8}\cdot\sqrt{2} = \sqrt{16} = 4\)
I believe you have it! Now, we know the "Adjacent". Now, off to your Pythagorean Theorem to find the Hypotenuse, EH.

- purplemexican

a^2+b^2=c^2\[4^2+3^2=5^2 \] and or 16+9=25 \[c=\sqrt{25}\] or c=5

- purplemexican

@tkhunny do you concur?

- tkhunny

We have the hypotenuse. Now, you can solve the problem. What is the requested cosine?

- purplemexican

indeed

- tkhunny

Eventually, you should also recognize "Pythagorean Triples" - at least 3-4-5 and 5-12-13. There are many others.

- purplemexican

i get cos(37)=0.765

- tkhunny

No one cares about those decimals or degrees.
\(\cos\left(\angle{FHE}\right) = \dfrac{4}{5}\) - Done!

- purplemexican

thank you so much i appreciate the help

- tkhunny

You got it. Keep up the good work. Start earlier so you CAN take a break. :-)

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