Answer: B
\( \sin(90^\circ + \theta) = \cos(120^\circ - 3\theta) \) Or, \( \sin(90^\circ + \theta) = \sin[90^\circ - (120^\circ - 3\theta)] \) (using the identity \( \cos x = \sin(90^\circ - x) \)) So, \( 90^\circ + \theta = 90^\circ - (120^\circ - 3\theta) \) Which simplifies to: \( 90^\circ + \theta = 90^\circ - 120^\circ + 3\theta \) Then, \( \theta - 3\theta = 90^\circ - 120^\circ - 90^\circ \) So, \( -2\theta = -120^\circ \) Therefore, \( \theta = 60^\circ \)
\( \sin(90^\circ + \theta) = \cos(120^\circ - 3\theta) \) Or, \( \sin(90^\circ + \theta) = \sin[90^\circ - (120^\circ - 3\theta)] \) (using the identity \( \cos x = \sin(90^\circ - x) \)) So, \( 90^\circ + \theta = 90^\circ - (120^\circ - 3\theta) \) Which simplifies to: \( 90^\circ + \theta = 90^\circ - 120^\circ + 3\theta \) Then, \( \theta - 3\theta = 90^\circ - 120^\circ - 90^\circ \) So, \( -2\theta = -120^\circ \) Therefore, \( \theta = 60^\circ \)