ALGEBRA 2/TRIG HELP PLEASE PLEASE HELP

- anonymous

ALGEBRA 2/TRIG HELP PLEASE PLEASE HELP

- jamiebookeater

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

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- anonymous

The point (2, 3) is on the terminal side of angle θ, in standard position. What are the values of sine, cosine, and tangent of θ? Make sure to show all work.

- anonymous

- anonymous

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- anonymous

- anonymous

- anonymous

|dw:1437061318167:dw|Ok, first thing you need to do is find what the hypotenuse is. It's a right triangle, so can use the Pythagorean Theorem to figure out what the (?) on the drawing is?

- anonymous

Is it 13?

- anonymous

Close! The hypotenuse is \(\sqrt{13}\) since the hypotenuse squared is 13.|dw:1437061696572:dw|

- anonymous

Okay :) what do I need to do next?

- anonymous

So, now, we need to remember what sine, cosine, and tangent represent on a right triangle. The mnemonic device is SOH CAH TOA.
S=Sine
C=Cosine
T=Tangent
O=Opposite
A=Adjacent
H=Hypotenuse
So, for example:
\[\sin(\theta)=\frac{\text{side opposite of }\theta}{\text{hypotenuse}}\]The other 2, cosine and tangent work the same way in the mnemonic device. So, can you tell me what sine would be using this triangle?

- anonymous

Sine would be 3/13, right? Which would make it 0.23076923076

- anonymous

Wait, I forgot to square 13

- anonymous

Remember, the hypotenuse is \(\sqrt{13}\), so \(\sin(\theta)=\frac{3}{\sqrt{13}}\)

- anonymous

would it be 0.83205029433

- anonymous

Nice. Using the mnemonic SOH CAH TOA, you can figure out what cosine and tangent represent. \[\cos(\theta)=\frac{\text{side adjacent to }\theta}{\text{hypotenuse}}\]\[\tan(\theta)=\frac{\text{side opposite to }\theta}{\text{side adjacent to }\theta}\]Can you figure out the rest?

- anonymous

Cosine would be 2/√13=0.55470019622 and tangent would be 3/2=1.5

- anonymous

Good job. I'd recommend for future reference, I'd just get it into your head that sine, cosine, and tangent are those ratios right there. It'll make your life easier as you go further into trig.

- anonymous

Thank you so much!! :) You were a great help.

- anonymous

Also, it didn't happen in this question, but remember, negatives stick with the calculation! So, if the point were in quadrant II (so the point would be (-2, 3)), then the x-value would be -2!! So, if you were given that, \(\sin(\theta)=\frac{3}{\sqrt{13}}\), \(\cos(\theta)=-\frac{2}{\sqrt{13}}\), and \(\tan(\theta)=-\frac{3}{2}\). See how the negative carried over?

- anonymous

Oh, I see

- anonymous

That's a good tip to remember; I almost always forget to carry the negative sign

- anonymous

Thanks again :)

- anonymous

np :)

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