Question

Consider the following figure:

$\sin B$ is $\frac{12}{m}$, m is

Answer 1

In $△ABC$, $AD\perp BC$ $AB=13$, $BD=5$, $DC=16$
Now, In $△ABD$,
$A{B}^2=A{D}^2+B{D}^2$
${13}^2=A{D}^2+5^2$
$A{D}^2=144$
$AD=12$
Now, in $△ADC$,
$A{C}^2=A{D}^2+C{D}^2$
$A{C}^2={12}^2+{16}^2$
$A{C}^2=144+256$
$AC=20$ cm
Now, $\sin B=\frac{P}{H}=\frac{AD}{AB}=\frac{12}{13}$