If B and Q are acute angles such that sin B = sin Q, then prove that B = Q.
Step-by-Step Solution
Key Concept: For acute angles (0° < θ < 90°) the sine function is strictly increasing; alternatively, use the identity \(\sin C-\sin D = 2\cos\frac{C+D}{2}\sin\frac{C-D}{2}\) and the fact that \(\cos\) of an acute angle is never zero.
1. Given \(\sin B = \sin Q\) with \(0^{\circ}
2. Subtract the two sides:
$$\sin B - \sin Q = 0.$$
3. Use the trigonometric identity:
$$\sin B - \sin Q = 2\cos\frac{B+Q}{2}\,\sin\frac{B-Q}{2}.$$
Hence
$$2\cos\frac{B+Q}{2}\,\sin\frac{B-Q}{2}=0.$$
4. For acute angles, \(\frac{B+Q}{2}\) is also acute (since the average of two acute angles is acute). Therefore \(\cos\frac{B+Q}{2}
eq 0\).
5. The product being zero forces the second factor to be zero:
$$\sin\frac{B-Q}{2}=0.$$
6. The sine of an angle is zero only when the angle is an integer multiple of \(180^{\circ}\). Because \(\frac{B-Q}{2}\) lies between \(-45^{\circ}\) and \(45^{\circ}\) (both acute), the only possible multiple is \(0^{\circ}\).
Hence
$$\frac{B-Q}{2}=0^{\circ}\quad\Rightarrow\quad B-Q=0^{\circ}.$$
7. Therefore, \(B = Q\).
8. Conclusion: If two acute angles have equal sines, the angles themselves are equal.
*Alternative reasoning (monotonicity):* Since \(\sin\theta\) is strictly increasing on \(0^{\circ}<\theta<90^{\circ}\), equality of sines directly implies equality of the angles.
Correct Answer: ∠B = ∠Q