baybay88
Help! what are the ratios for sin X and cos X?
drawing below
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amistre64
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sin = over/hyp
cos = next.to/hyp
baybay88
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|dw:1370976541873:dw|
baybay88
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so how do i write it with those numbers?
amistre64
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memory tricks may be:
theres a SIGN, OVER there
COSy up NEXT.TO someone you like
baybay88
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haha thats good!! i forgot which one was the hyp though...
amistre64
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|dw:1370976658919:dw|
amistre64
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hyp is the side that is neither over or next.to :) its the slanted part of the rt tri
baybay88
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oh so then to 5\[5 \over 119?\]
baybay88
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sorry math isnt the best for me lol
amistre64
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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
baybay88
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ok so then it would be 119/12 for sin and 5/12 for cos?
amistre64
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yes, but dont forget to notate that 119 properly: \(\sqrt{119}\)
baybay88
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ok so if i did it write then the sin would be 130.904?
baybay88
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*right
baybay88
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and cos would be 0.416?
amistre64
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your cosine is fine
you hit some wrong buttons on your calculator for the sin tho
baybay88
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oh really?oh yeah I did that wrong. Thank you!!
amistre64
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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 :)
amistre64
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good luck
baybay88
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thank you!