Trigonometry (11th Edition) Clone

Published by Pearson
ISBN 10: 978-0-13-421743-7
ISBN 13: 978-0-13421-743-7

Chapter 2 - Test - Page 97: 4a



Work Step by Step

Both angles are In the interval $0^{\mathrm{o}}\leq\theta\leq 90^{\mathrm{o}}$, (quadrant I) . where sine increases from $0$ to $1$ (as angles get larger, their sine gets larger ) since $24^{\mathrm{o}} < 48^{\mathrm{o}}$, it follows that $\sin 24^{\mathrm{o}} < \sin 48^{\mathrm{o}}$ so, the statement is true.
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