If cos theta = -3/5 in quadrant 2, what is sin theta?

Step-by-step explanation:

if cos theta = -3/5 in quadrant 2, find out sin(theta)

cos is negative in second and third quadrant

Here cos is in second quadrant.

Sin is positive in second quadrant .

adjacent side= 3 , hypotenuse = 5

Lets use Pythagorean theorem

Subtract 9 from both sides

b=4

opposite side = 4

hello : sin(θ) = 4/25

Step-by-step explanation:

you know : cos²(θ) + sin²(θ) =1   and : cos(θ) = - 3/5

(-3/5)² + sin²(θ) =1

9/25 + sin²(θ) =1

sin²(θ) =1 -9/25

sin²(θ) = 16/25

sin(θ) = 4/25 or sin(θ) = - 4/25

in quadrant 2 :  sin(θ) > 0

so : sin(θ) = 4/25

Just draw 3 non parallel lines. These paths will form a triangle with vertices A, B, C. So you know that the angles have to add up to 180.  

Step-by-step explanation: I HAD A DIFFICULT TIME DOING THIS ON PLATO I'LL COME BACK WITH AN UPDATE TO SEE IF ITS CORRECT. ( I HOPE I HELP FOR NOW)