+194 In right triangle ABC with angle B = 90˚ , we have 2sinA = 3cosA . What is sinA?
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
+9673 "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) = \(\dfrac{3\sqrt{13}}{13}\)
