I need help finding sinA if

cosA=(-3/5) and A is greater than 90 degrees, but less than 180 degrees.

Draw the quadrant diagram. In the upper left quadrant, greater then ninety degrees, but less than 180, cosine is neg, and sine is positive.

Draw a three, four, five right triangle.....Look at the angle where cos is 4/5, what is the sine?

I am given a triangle where i have to find the angle across from the right angle and i do not know how to find this angle with given lengths. please helpppp

To find the sine of an angle, we can use the Pythagorean identity:

sin^2(A) + cos^2(A) = 1

We know that cos(A) = -3/5, so we can substitute this value into the equation:

sin^2(A) + (-3/5)^2 = 1
sin^2(A) + 9/25 = 1

Next, we can move the constant term to the other side of the equation:

sin^2(A) = 1 - 9/25
sin^2(A) = 25/25 - 9/25
sin^2(A) = 16/25

Finally, we take the square root of both sides to solve for sin(A):

sin(A) = ±sqrt(16/25)
sin(A) = ±(4/5)

Since A is between 90 and 180 degrees, in the upper left quadrant, sine is positive. Therefore, sin(A) = 4/5.