Question

two right triangles, △abd and △acd, were created from △abc by constructing ad ⊥ cb. follow these steps to prove that $\frac{sin(b)}{b}=\frac{sin(c)}{c}$. sin(b) = $\frac{h}{c}$ becomes c sin(b)=h. 3. for △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 b sin(c)=h. 5. by the property, we can set the expressions equal to each other. c sin(b)=b sin(c)

Explanation:

Step1: Recall sine - definition in right - triangle

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

Step2: Use transitive property

Since $c\sin(B) = h$ and $b\sin(C)=h$, by the transitive property (if $a = c$ and $b = c$, then $a = b$), we can set $c\sin(B)=b\sin(C)$. Then, dividing both sides by $bc$ gives $\frac{\sin(B)}{b}=\frac{\sin(C)}{c}$.

Answer:

Transitive