#### Answer

true

#### 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.