finding a value sin 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...
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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.
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