Question

help please

Answer 1

Please provide the Right-Triangle definition of the Cosine.

Answer 2

everywhere

Answer 3

the most i can tell you is what im looking for

Answer 4

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

Answer 5

do i know what?

Answer 6

i have been given what you see

Answer 7

and 4 answer choices

Answer 8

if i knew earlier today i dont remember now

Answer 9

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.

Answer 10

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

Answer 11

im in summerschool

Answer 12

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

Answer 13

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

Answer 14

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

Answer 15

Did you read it?

Answer 16

yes it make sence

Answer 17

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

Answer 18

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

Answer 19

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

Answer 20

\(\cos(\angle FHE) = \dfrac{Adjacent}{Hypotenuse}\)

You know neither the Adjacent Side (FH) nor the Hypotenuse (EH). How shall we find them?

Answer 21

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

Answer 22

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

Answer 23

yes it would

Answer 24

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

Answer 25

leg FH is 4 right

Answer 26

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.

Answer 27

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

Answer 28

@tkhunny do you concur?

Answer 29

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

Answer 30

indeed

Answer 31

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

Answer 32

i get cos(37)=0.765

Answer 33

No one cares about those decimals or degrees.

\(\cos\left(\angle{FHE}\right) = \dfrac{4}{5}\) - Done!

Answer 34

thank you so much i appreciate the help

Answer 35

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