Question

two right triangles, △abd and △acd, were created from △abc by constructing ad ⊥ cb. follow these steps to prove that $\frac{sin(b)}{b}=\frac{sin(c)}{c}$. 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 $csin(b)=h$. 3. for △acd, $sin(c)=$

Explanation:

Step1: Recall sine - definition in right - triangle

In right - triangle $\triangle ACD$, the sine of an angle in a right - triangle is defined as the ratio of the opposite side to the hypotenuse. For $\angle C$ in $\triangle ACD$, the opposite side to $\angle C$ is $h$ and the hypotenuse is $b$.

Step2: Write the sine formula

So, $\sin(C)=\frac{h}{b}$.

Answer:

$\frac{h}{b}$