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prisoha [69]
3 years ago
14

The point (−3, 1) 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.
Mathematics
2 answers:
lara31 [8.8K]3 years ago
8 0

ANSWER

See below

EXPLANATION

The given point (-3,1) lies in the second quadrant.

In this quadrant only sine is positive.

The length of the hypotenuse formed by the right angle triangle is

{h}^{2}  =  {3}^{2}  +   {1}^{2}

{h}^{2}  =  9 +  1

{h}^{2}  = 10

h =  \sqrt{10}

The side opposite to θ, is 1 units.

The adjacent side is 3 units.

\sin( \theta)  =  \frac{opposite}{hypotenuse}

\sin( \theta)  =  \frac{1}{ \sqrt{10} }  =  \frac{ \sqrt{10} }{10}

\cos( \theta)  =  -  \frac{adjacent}{hypotenuse}

\cos( \theta)  = -   \frac{3}{ \sqrt{10} }  =  -  \frac{3 \sqrt{10} }{10}

\tan( \theta)  =  -  \frac{opposite}{adjacent}

\tan( \theta)  = -   \frac{1}{3}

ELEN [110]3 years ago
6 0

Answer:

The angle is in the second quadrant  so only the sine is positive . 

Length of the hypotenuse = sqrt (-1^2 + 3^2)  = sqrt10

sine  =  1/sqrt10 =  0.3162 to 4 dec places

tangent = 1 / -3 =  -0.3333  to 4 d p's

cosine = -3/sqrt10 = -0.9487 to 4 d, p's.

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f(-3) = -2

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Step-by-step explanation:

(Whole question:

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