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
[tex] {h}^{2} = {3}^{2} + {1}^{2} [/tex]
[tex] {h}^{2} = 9 + 1[/tex]
[tex]{h}^{2} = 10[/tex]
[tex]h = \sqrt{10} [/tex]
The side opposite to θ, is 1 units.
The adjacent side is 3 units.
[tex] \sin( \theta) = \frac{opposite}{hypotenuse} [/tex]
[tex]\sin( \theta) = \frac{1}{ \sqrt{10} } = \frac{ \sqrt{10} }{10} [/tex]
[tex]\cos( \theta) = - \frac{adjacent}{hypotenuse} [/tex]
[tex]\cos( \theta) = - \frac{3}{ \sqrt{10} } = - \frac{3 \sqrt{10} }{10} [/tex]
[tex] \tan( \theta) = - \frac{opposite}{adjacent} [/tex]
[tex]\tan( \theta) = - \frac{1}{3} [/tex]
