Question

If \(\sin(30^\circ + \theta) = \cos 40^\circ\), then the value of \(\theta\) is :

• A. \(40^\circ\)
• B. \(30^\circ\)
• C. \(20^\circ\)
• D. \(50^\circ\)

Hint:

Use the complementary angle identity: \(\cos A = \sin(90^\circ - A)\). 1. Given: \(\sin(30^\circ + \theta) = \cos 40^\circ\). 2. Rewrite \(\cos 40^\circ\) as \(\sin(90^\circ - 40^\circ) = \sin 50^\circ\). 3. So, \(\sin(30^\circ + \theta) = \sin 50^\circ\). 4. Equate the angles: \(30^\circ + \theta = 50^\circ\). 5. Solve for \(\theta\): \(\theta = 50^\circ - 30^\circ = 20^\circ\).

Answer 1

The Correct Option is C

Concept: This problem uses the complementary angle identity for sine and cosine, which states that \(\cos \phi = \sin(90^\circ - \phi)\) or \(\sin \phi = \cos(90^\circ - \phi)\). Step 1: Use the complementary angle identity to express \(\cos 40^\circ\) in terms of sine We know that \(\cos \phi = \sin(90^\circ - \phi)\). Let \(\phi = 40^\circ\). So, \(\cos 40^\circ = \sin(90^\circ - 40^\circ) = \sin 50^\circ\). Step 2: Substitute this into the given equation The given equation is \(\sin(30^\circ + \theta) = \cos 40^\circ\). Substituting \(\cos 40^\circ = \sin 50^\circ\), we get: \[ \sin(30^\circ + \theta) = \sin 50^\circ \] Step 3: Solve for \(\theta\) If \(\sin A = \sin B\), then one possible solution (for acute angles, which is usually the context here unless specified) is \(A = B\). So, we can equate the angles: \[ 30^\circ + \theta = 50^\circ \] Subtract \(30^\circ\) from both sides: \[ \theta = 50^\circ - 30^\circ \] \[ \theta = 20^\circ \] Alternative approach using \(\sin \phi = \cos(90^\circ - \phi)\) Alternatively, we could convert \(\sin(30^\circ + \theta)\) to cosine form: \(\sin(30^\circ + \theta) = \cos(90^\circ - (30^\circ + \theta)) = \cos(90^\circ - 30^\circ - \theta) = \cos(60^\circ - \theta)\). So the equation becomes: \(\cos(60^\circ - \theta) = \cos 40^\circ\). Equating the angles: \(60^\circ - \theta = 40^\circ\) \(60^\circ - 40^\circ = \theta\) \(\theta = 20^\circ\). The value of \(\theta\) is \(20^\circ\). This matches option (3).