Question

If \( \sin 3\theta = \cos(\theta - 6^\circ) \), where \( (3\theta) \) and \( (\theta - 6^\circ) \) are both acute angles then the value of \( \theta \) is :

• A. \(18^\circ\)
• B. \(24^\circ\)
• C. \(36^\circ\)
• D. \(30^\circ\)

Hint:

Use the identity: If \(\sin A = \cos B\), then \(A+B = 90^\circ\) (for acute A, B). Here, \(A = 3\theta\) and \(B = \theta - 6^\circ\). So, \(3\theta + (\theta - 6^\circ) = 90^\circ\). \(4\theta - 6^\circ = 90^\circ\). \(4\theta = 96^\circ\). \(\theta = 24^\circ\). Always check if the resulting angles \(3\theta\) and \(\theta-6^\circ\) are indeed acute. \(3(24^\circ) = 72^\circ\) (acute). \(24^\circ - 6^\circ = 18^\circ\) (acute).

Answer 1

The Correct Option is B

Concept: This problem uses the complementary angle identity for sine and cosine: \(\sin x = \cos(90^\circ - x)\) or \(\cos y = \sin(90^\circ - y)\). If \(\sin A = \cos B\), and A and B are acute angles, then \(A + B = 90^\circ\). Step 1: Apply the complementary angle relationship We are given \( \sin 3\theta = \cos(\theta - 6^\circ) \). Since \(3\theta\) and \((\theta - 6^\circ)\) are acute angles, we can use the property that if \(\sin A = \cos B\), then \(A + B = 90^\circ\). Let \(A = 3\theta\) and \(B = \theta - 6^\circ\). So, \[ A + B = 90^\circ \] \[ 3\theta + (\theta - 6^\circ) = 90^\circ \] Step 2: Solve for \(\theta\) \[ 3\theta + \theta - 6^\circ = 90^\circ \] \[ 4\theta - 6^\circ = 90^\circ \] Add \(6^\circ\) to both sides: \[ 4\theta = 90^\circ + 6^\circ \] \[ 4\theta = 96^\circ \] Divide by 4: \[ \theta = \frac{96^\circ}{4} \] \[ \theta = 24^\circ \] Step 3: Verify that the angles are acute If \(\theta = 24^\circ\):
\(3\theta = 3 \times 24^\circ = 72^\circ\). This is an acute angle (between \(0^\circ\) and \(90^\circ\)).
\(\theta - 6^\circ = 24^\circ - 6^\circ = 18^\circ\). This is also an acute angle. The conditions are satisfied. Therefore, the value of \(\theta\) is \(24^\circ\).