## anonymous one year ago help please

1. anonymous

2. tkhunny

Please provide the Right-Triangle definition of the Cosine.

3. anonymous

everywhere

4. anonymous

the most i can tell you is what im looking for

5. tkhunny

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

6. anonymous

do i know what?

7. anonymous

i have been given what you see

8. anonymous

9. anonymous

if i knew earlier today i dont remember now

10. 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.

11. anonymous

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

12. anonymous

im in summerschool

13. anonymous

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

14. anonymous

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

15. tkhunny

16. anonymous

yes it make sence

17. tkhunny

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

18. anonymous

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

19. anonymous

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

20. tkhunny

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

21. anonymous

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

22. tkhunny

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

23. anonymous

yes it would

24. tkhunny

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

25. anonymous

leg FH is 4 right

26. 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.

27. anonymous

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

28. anonymous

@tkhunny do you concur?

29. tkhunny

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

30. anonymous

indeed

31. tkhunny

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

32. anonymous

i get cos(37)=0.765

33. tkhunny

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

34. anonymous

thank you so much i appreciate the help

35. tkhunny

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