Question

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

1. for △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

Explanation:

Step1: Recall sine - definition in right - triangle

In right - triangle $\triangle ABD$, $\sin(B)=\frac{h}{c}$ (where $h$ is the length of the altitude $AD$ and $c$ is the length of side $AB$).

Step2: Multiply both sides by $c$

Multiply $\sin(B)=\frac{h}{c}$ by $c$. We get $c\sin(B)=h$.

Step3: Consider $\triangle ACD$

In right - triangle $\triangle ACD$, $\sin(C)=\frac{h}{b}$, so $h = b\sin(C)$.

Step4: Equate the two expressions for $h$

Since $c\sin(B)=h$ and $h = b\sin(C)$, we can set $c\sin(B)=b\sin(C)$.

Step5: Rearrange the equation

Divide both sides of $c\sin(B)=b\sin(C)$ by $bc$. We obtain $\frac{\sin(B)}{b}=\frac{\sin(C)}{c}$.

Answer:

$\frac{\sin(B)}{b}=\frac{\sin(C)}{c}$