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QUESTION 5 \cdot 1 POINT Given that $\cot(\theta) = \frac{9}{40}$ and $\theta$ is in Quadrant III, find $\sin(\theta)$ and $\tan(\theta)$. Give exact answers in the form of a fraction. Provide your answer below:

          QUESTION 5 \cdot 1 POINT
Given that $\cot(\theta) = \frac{9}{40}$ and $\theta$ is in Quadrant III, find $\sin(\theta)$ and $\tan(\theta)$.
Give exact answers in the form of a fraction.
Provide your answer below:
        
QUESTION 5 ·1 POINT
Given that cot(θ) = (9)/(40) and θ is in Quadrant III, find sin(θ) and tan(θ).
Give exact answers in the form of a fraction.
Provide your answer below:

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Precalculus with Limits
Precalculus with Limits
Ron Larson 2nd Edition
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Content attribution QUESTION 5 - 1 POINT Given that cot(θ) = 9/40 and θ is in Quadrant III, find sin(θ) and tan(θ). Give exact answers in the form of a fraction. Provide your answer below: ~ 47 days 5 FEEDBACK Content attribution QUESTION 5 - 1 POINT Given that cot(θ) = 9/40 and θ is in Quadrant III, find sin(θ) and tan(θ). Give exact answers in the form of a fraction. Provide your answer below:
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Transcript

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00:01 Okay, so we're given the tangent of theta equals four thirds, and we know that we're working in quadrant three.
00:07 So the first thing we're going to do is we're going to draw our triangle here.
00:12 We know that tangent is the opposite side over the adjacent side.
00:18 So our triangle will look like this, and i'll call this leg c because we're not sure what its length is.
00:26 And we can apply the pythagorean.
00:28 4 squared plus 3 squared equals c squared if you add this out we'll get 25 equals c squared so our c has to be 5 so this equals 5 so now we should be able to fill in the rest of our chart sign is opposite over hypotenuse but remember we're in quadrant 3 so our only angle measure that will be positive is tangent and co -tangent.
01:02 All others will be negative.
01:03 So this will be the negative opposite over hypotenuse...
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