Question

given: △acz and △cbz are right triangles. prove: $\frac{sin(a)}{a}=\frac{sin(b)}{b}$. statements: △acz and △cbz are right triangles (given), $bsin(a)=h$, $asin(b)=h$ (multiplication property of equality), $bsin(a)=asin(b)$ (transitive property of equality), $\frac{sin(a)}{a}=\frac{sin(b)}{b}$ (division property of equality). what is the missing step of the proof? a. $\frac{a}{b}=\frac{sin(a)}{sin(b)}$; corresponding sides are proportional in similar triangles. b. $\frac{a}{b}=\frac{sin(b)}{sin(a)}$; corresponding sides are proportional in similar triangles. c. $sin(a)=\frac{h}{b}$, $sin(b)=\frac{h}{a}$; the definition of the sine ratio. d. $sin(b)=\frac{h}{b}$, $sin(a)=\frac{h}{a}$; the definition of the sine ratio.

Explanation:

Step1: Recall sine - ratio in right - triangles

In right - triangle $\triangle ACZ$, $\sin(A)=\frac{h}{b}$, so $b\sin(A)=h$. In right - triangle $\triangle CBZ$, $\sin(B)=\frac{h}{a}$, so $a\sin(B)=h$.

Step2: Use the transitive property of equality

Since $b\sin(A) = h$ and $a\sin(B)=h$, by the transitive property of equality, $b\sin(A)=a\sin(B)$.

Step3: Apply the division property of equality

Divide both sides of the equation $b\sin(A)=a\sin(B)$ by $ab$: $\frac{\sin(A)}{a}=\frac{\sin(B)}{b}$.

Answer:

A. $\frac{\sin(A)}{a}=\frac{\sin(B)}{b}$; corresponding sides are proportional in similar triangles