The angles \(3A\) and \(2A-15^\circ\) are both acute, and \(\sin 3A=\cos(2A-15^\circ)\). Find \(A\).
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(21^\circ\)
Since \(\sin 3A=\cos(90^\circ-3A)\), the condition says that two acute angles have the same cosine. Two acute angles with equal cosines are equal, so \(90^\circ-3A=2A-15^\circ\).
\(\cos(90^\circ-3A)=\cos(2A-15^\circ)\), so \(90^\circ-3A=2A-15^\circ\)
\(5A=105^\circ\), so \(A=21^\circ\)
Check: \(3A=63^\circ\) and \(2A-15^\circ=27^\circ\) are both acute and add up to \(90^\circ\)