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Mathematics 11 Online
OpenStudy (anonymous):

PLEASE HELP! find the value of the sine for angle A?

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?

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 !

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

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

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.

OpenStudy (anonymous):

so sqrt5/3! thank you so much for helping me

OpenStudy (math&ing001):

You're very welcome !

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