PLEASE HELP! find the value of the sine for angle A?
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OpenStudy (anonymous):
OpenStudy (math&ing001):
Do you know how we find the sinus of an angle in general ?
OpenStudy (anonymous):
sinx=o/h ?
OpenStudy (math&ing001):
Yeah ! Can you apply it here
OpenStudy (anonymous):
i found the hypotenuse and got 14.42 is that right? and then aren't i just supposed to divide the opposite side by the hypotenuse? aka 14.42?
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OpenStudy (math&ing001):
Actually, here, the hypotenuse is already given and it's 12, but you'll need to find the opposite using the Pythagorean theorem. And then yes it's just a simple division.
OpenStudy (anonymous):
when i do the Pythagorean theorem i get 14.42
OpenStudy (math&ing001):
Not possible, the opposite is always less than the hypotenuse. |dw:1416586430765:dw| \[a=\sqrt{12^{2}-8^{2}}\]
OpenStudy (anonymous):
8.9?
OpenStudy (math&ing001):
Correct !
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OpenStudy (math&ing001):
And then use it to find the sin.
OpenStudy (anonymous):
i got 1.34///
OpenStudy (anonymous):
but i dont think thats right?
OpenStudy (math&ing001):
How did you get that?
OpenStudy (anonymous):
i divided 12 and 8.9
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OpenStudy (math&ing001):
Yeah but it's opposite/hypo
OpenStudy (math&ing001):
opposite = 8.9 & hypo = 12
OpenStudy (anonymous):
wait so im confused? i just did it backwards then ?
OpenStudy (math&ing001):
Yep !
OpenStudy (anonymous):
0.74 then
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OpenStudy (math&ing001):
Correct !
OpenStudy (anonymous):
how do i get my answer into fraction form then?
OpenStudy (math&ing001):
Then we don't evaluate :)
\[\sin(A) = \frac{ \sqrt{12^{2}-8^{2}} }{ 12 }=\frac{ \sqrt{80} }{ 12 }\] You can simplify it more if you like.