Question

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) =$2. multiply both sides of the equation by $c$ and simplify. the equation $\sin(b)=\frac{h}{c}$ becomes $c\sin(b)=h$.3. for $\triangle acd$, $\sin(c) =$

Explanation:

Step1: Define sin(C) in △ACD

In right triangle $\triangle ACD$, $\sin(\text{angle}) = \frac{\text{opposite side}}{\text{hypotenuse}}$. For $\angle C$, the opposite side is $h$, and the hypotenuse is $b$.
$\sin(C) = \frac{h}{b}$

Answer:

$\frac{h}{b}$