## baybay88 2 years ago Help! what are the ratios for sin X and cos X? drawing below

1. amistre64

sin = over/hyp cos = next.to/hyp

2. baybay88

|dw:1370976541873:dw|

3. baybay88

so how do i write it with those numbers?

4. amistre64

memory tricks may be: theres a SIGN, OVER there COSy up NEXT.TO someone you like

5. baybay88

haha thats good!! i forgot which one was the hyp though...

6. amistre64

|dw:1370976658919:dw|

7. amistre64

hyp is the side that is neither over or next.to :) its the slanted part of the rt tri

8. baybay88

oh so then to 5$5 \over 119?$

9. baybay88

sorry math isnt the best for me lol

10. amistre64

if we observe the X angle: over = sqrt(119) next.to = 5 that leaves hyp=12 sin X = over/hyp cos X = next.to/hyp its just a matter of filling in the proper parts

11. baybay88

ok so then it would be 119/12 for sin and 5/12 for cos?

12. amistre64

yes, but dont forget to notate that 119 properly: $$\sqrt{119}$$

13. baybay88

ok so if i did it write then the sin would be 130.904?

14. baybay88

*right

15. baybay88

and cos would be 0.416?

16. amistre64

your cosine is fine you hit some wrong buttons on your calculator for the sin tho

17. amistre64
18. baybay88

oh really?oh yeah I did that wrong. Thank you!!

19. amistre64

a good dbl chk is that sin^2 + cos^2 = 1 sin^2 + (5/12)^2 = 1 sin^2 + 25/144 = 1 sin^2 = 1 - 25/144 sin^2 = 119/144 sin = sqrt(119)/12 but then youd still ahv eto hit the right buttons :)

20. amistre64

good luck

21. baybay88

thank you!