SOLUTION: solve the equation 3cos A + 2sin A = 2.8 for 0 degrees<A<360 degrees
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Question 845424: solve the equation 3cos A + 2sin A = 2.8 for 0 degrees
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
solve the equation 3cos A + 2sin A = 2.8
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3cos(A) + 2sqrt(1-cos^2(A)) = 2.8
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2sqrt(1-cos^2(A)) = 2.8-3cos(A)
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Square both sides to get:
4-4cos^2(A)) = 2.8^2 - 16.8cosA + 9cos^2(A)
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13cos^2(A) - 16.8cos(A) + 3.84 = 0
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cos(A) = [16.8 +- sqrt(16.8^2 -4*9*3.84)]/26
cos(A) = [16.8+-12]/26
cos(A) = 4.8/26 = 0.1846
A = arccos(0.1846) = 79.36 degrees
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Cheers,
Stan H.
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