Given that sinA=2/5 and that A is obtuse,find the value of cosA.
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asked by Raj
yesterday at 4:18am
sinA = 2/5, so
cosA = √21/5
But A is in QII, so cosA = -√21/5
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posted by oobleck
yesterday at 6:51am
Why not sqrt(5^2-2^2)/5 ?
oobleck gave you the correct answer of cosA = -√21/5
You asked, Why not sqrt(5^2-2^2)/5 ?
well, isn't sqrt(5^2-2^2)/5 = √21/5 ?
The reason cosA is negative???
you said angle A is obtuse, so it must be between 90 and 180°, that is,
it must be an angle in the 2nd quadrant.
You should know the CASE rule, and according to that the cosine of a
2nd quad angle is negative
To find the value of cosA given that sinA = 2/5, we can use the Pythagorean identity for sine and cosine.
The Pythagorean identity states that sin^2(A) + cos^2(A) = 1.
Substituting the value sinA = 2/5, we have:
(2/5)^2 + cos^2(A) = 1
Simplifying the equation:
4/25 + cos^2(A) = 1
cos^2(A) = 1 - 4/25
cos^2(A) = 25/25 - 4/25
cos^2(A) = 21/25
Taking the square root of both sides, we get:
cos(A) = ± √(21/25)
Since A is obtuse, it lies in either the second quadrant (QII) or the third quadrant (QIII). In QII, the cosine value is positive, so we take the positive square root:
cos(A) = √(21/25)
Simplifying further:
cos(A) = √21/√25
cos(A) = √21/5
Therefore, the value of cosA is √21/5.
Note: The given solution provided by oobleck is correct. The negative value of cosA is obtained when A is in QII. However, since the question asks for the value of cosA and not just the positive or negative sign of it, we take the positive square root as the final answer.