Question

having a hard time with sin, cos, and tan with that zero like thing next to it (I don't remember the name). Help?

Answer 1

\(\sin(\theta)\) ?

Answer 2

Yes, that.

Answer 3

I'm needing to find the tan, sin, and cos for that triangle and it's online course so i don't really have any help.

Answer 4

sine is opposite over hypotenuse
you find the opposite side by pythagoras

Answer 5

lets call that side \(b\) then pythagoras tells you that
\[10^2+b^2=12^2\]
\[b^2=144-100\]
\[b^2=44\]
\[b=\sqrt{44}=2\sqrt{11}\]

Answer 6

I get to the point in solving the problem where you do the square root of blank = A and then i get lost.

Answer 7

I don't understand how you got that or what its for.

Answer 8

now it is a matter of knowing the ratios
\[\sin(\theta)=\frac{2\sqrt{11}}{12}=\frac{\sqrt{11}}{6}\]
\[\cos(\theta)=\frac{10}{12}=\frac{5}{6}\] \[\tan(\theta )=\frac{2\sqrt{11}}{10}=\frac{\sqrt{11}}{5}\]

Answer 9

Oh! so you simplified ?

Answer 10

I'm sorry I'm terrible when it comes to math..

Answer 11

\[a^2+b^2=c^2\]
if you know any two of the numbers you can find the third

Answer 12

you get
\[10^2+b^2=12^2\]
\[100+b^2=144\]
\[b^2=44\]
\[b=\sqrt{44}\]

Answer 13

I understand the a2+b2=c2 but I get through like maybe 3 steps and et confused.

Answer 14

are those steps ok?

Answer 15

Yes. It's right after that last step I don't know what to do.

Answer 16

I have a dvd that explains stuff for these problems and he's saying you multiply something to get some letter and im just not sure whats going on.

Answer 17

\[44=4\times 11\]
\[\sqrt{44}=\sqrt{4\times 11}=\sqrt{4}\times \sqrt{11}=2\sqrt{11}\]

Answer 18

it is just another way to write \(\sqrt{44}\) so you can figure out how to cancel when you take the ratios

Answer 19

Ok, so you take the number (44), then do the square root of 4 and 11 since they make 44 when multiplied then how did you get 2 sqrt 11?

Answer 20

i know that the square root of 4 is 2 so i replace \(\sqrt{4}\) by \(2\)
i don't know what the square root of 11 is, so i just leave it as \(\sqrt{11}\)

Answer 21

Oh! Okay. This is making more sense now. Thank you so much for your time/help!