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

Explanation:

Step1: Multiply both sides by $b$

$\sin(C) \times b = \frac{h}{b} \times b$

Step2: Simplify the equation

$b\sin(C) = h$

Answer:

$b\sin(C) = h$