# finding a valuesin A=? cos A = -(3/5) 90 degrees < A < 180 degrees

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Given that cos(A) = -3/5

We need to find sin(A) suchthat A is in the second quadrant.

First let us calculate the value of sin(A).

We know that sin^2 A + cos^2 A = 1

==> sin^2 A = 1- cos^2 A

==> sin^2 A = 1- (-3/5)^2

= 1- 9/25

= 16/25

==> sin(A) = +- 4/5

But since A is in the 2nd quadrant, then we know that the sine is positive.

**==> sin(A) = 4/5**

The angle A is in the 2nd quadrant, so the value of the function sine is positive.

We'll apply the fundamental formula of trigonometry:

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

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

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

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

(sin A)^2 = 16/25

**sin A = 4/5**

**We'll keep only the positive value, since we've established that the value of the sine function is positive in the 2nd quadrant.**