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

2 Answers | Add Yours

hala718's profile pic

hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted on

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

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

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.

We’ve answered 318,988 questions. We can answer yours, too.

Ask a question