Question

two right triangles, △abd and △acd, were created from △abc by constructing ad ⊥ cb. follow these steps to prove that $\frac{sin(b)}{b}=\frac{sin(c)}{c}$. multiplication by b and simplify. the equation. substitution can set the expressions equal to each other. $csin(b)=bsin(c)$ check

Explanation:

Step1: Define sine in right - triangles

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

Step2: Set the two expressions for $h$ equal

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

Step3: Rearrange the equation

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

Answer:

The proof is completed as shown above, and $\frac{\sin(B)}{b}=\frac{\sin(C)}{c}$ is proved.