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In right triangle ABC with angle B = 90˚ , we have 2sinA = 3cosA . What is sinA?

 Jul 14, 2018
 #1
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Solve for A:
2 sin(A) = 3 cos(A)

Divide both sides by cos(A):
2 tan(A) = 3

Divide both sides by 2:
tan(A) = 3/2

Take the inverse tangent of both sides:
Arctan(3/2) =A =56.31 degrees.

Sin(A) = sin(56.31) =0.832

 Jul 14, 2018
edited by Guest  Jul 14, 2018
 #2
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"In right triangle ABC with angle B = 90˚" implies that 0˚ < A < 90˚.

2 sin A = 3 cos A

2 tan A = 3

tan A = 3/2

A = arctan(3/2) = arccot(2/3)

sin A = sin(arccot(2/3)) = 1/sqrt(csc^2(arccot(2/3)) = 1/sqrt(1+cot^2(arccot(2/3))) = 1/sqrt(1+4/9) = \(\dfrac{3\sqrt{13}}{13}\)

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 Jul 15, 2018

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