## Trigonometry (11th Edition) Clone

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.