Question

the law of sines
two right triangles, △abd and △acd, were created from △abc by constructing $overline{ad} \perp \overline{cb}$.
follow these steps to prove that $\frac{\sin(b)}{b} = \frac{\sin(c)}{c}$.
\checkmark 1. for △abd, $\sin(b) = h/c$

1. multiply both sides of the equation by $c$ and simplify. the equation $\sin(b) = \frac{h}{c}$ becomes dropdown with options: $c \sin(b) = h$, $h \sin(b) = h$.

diagram: triangle $abc$ with $ad \perp cb$, labeled with side $b$ (ac), side $c$ (ab), height $h$ (ad), points $c$, $d$, $b$, $a$.

Explanation:

Step1: Start with the given equation

We have the equation $\sin(B) = \frac{h}{c}$.

Step2: Multiply both sides by \( c \)

To eliminate the denominator \( c \) on the right - hand side, we multiply both sides of the equation $\sin(B)=\frac{h}{c}$ by \( c \). Using the multiplication property of equality (if \( a = b \), then \( a\times c=b\times c \)), we get:
\( c\times\sin(B)=c\times\frac{h}{c} \)
Simplifying the right - hand side, \( c\times\frac{h}{c}=h \) (since \( c
eq0 \), we can cancel out the \( c \) terms). So the equation becomes \( c\sin(B)=h \).

Answer:

\( c\sin(B) = h \)