Question

If \(\sin 3A = \cos (A - 26^{\circ})\), where 3A is an acute angle, find the value of A.

• A. \(29^{\circ}\)
• B. \(61^{\circ}\)
• C. \(51^{\circ}\)
• D. \(39^{\circ}\)

Hint:

For any equation of the form \(\sin X = \cos Y\) where both are acute angles, the angles must be complementary:
\[ X + Y = 90^{\circ} \]
Here, \(3A + (A - 26^{\circ}) = 90^{\circ} \implies 4A - 26^{\circ} = 90^{\circ} \implies 4A = 116^{\circ} \implies A = 29^{\circ}\).
This shortcut saves steps in matching co-functions.

Answer 1

The Correct Option is A

Step 1: Apply the co-function identity to the right side this time.
We are given $\sin 3A = \cos(A - 26^{\circ})$. Instead of converting the sine on the left, let us convert the cosine on the right using $\cos\alpha = \sin(90^{\circ} - \alpha)$.
Step 2: Rewrite the right hand side. \[ \cos(A - 26^{\circ}) = \sin\big(90^{\circ} - (A - 26^{\circ})\big) = \sin(116^{\circ} - A) \]
Step 3: Now both sides are sine functions, so equate the angles. \[ \sin 3A = \sin(116^{\circ} - A) \implies 3A = 116^{\circ} - A \]
Step 4: Solve for A. \[ 4A = 116^{\circ} \] \[ A = 29^{\circ} \] \[ \boxed{29^{\circ}} \]