Question

proving the law of sines
two right triangles, $\triangle abd$ and $\triangle acd$, were created from $\triangle abc$ by constructing $\overline{ad} \perp \overline{cb}$.
follow these steps to prove that $\frac{\sin(b)}{b} = \frac{\sin(c)}{c}$.

1. for $\triangle abd$, $\sin(b) = $

Explanation:

Step1: Recall sine definition for right triangles

In a right triangle, $\sin(\text{angle}) = \frac{\text{opposite side}}{\text{hypotenuse}}$.

Step2: Identify sides for $\angle B$

In $\triangle ABD$, the right angle is at $D$. The side opposite $\angle B$ is $h$, and the hypotenuse is $c$.
<Expression>
$\sin(B) = \frac{h}{c}$
</Expression>

Answer:

$\frac{h}{c}$