Question

bret kamala
(sin ( c ) = \frac { c } { h }) (sin ( c ) = \frac { h } { a })
(sin ( a ) = \frac { a } { h }) (sin ( a ) = \frac { h } { c })
(h = \frac { a } { sin ( a ) }) (a sin ( c ) = h)
(h = \frac { c } { sin ( c ) }) (c sin ( a ) = h)
(\frac { a } { sin ( a ) } = \frac { c } { sin ( c ) }) (a sin ( c ) = c sin ( a ))
(\frac { sin ( c ) } { c } = \frac { sin ( a ) } { a }) (\frac { sin ( c ) } { c } = \frac { sin ( a ) } { a })
the proof was correctly completed by
the next step in the proof is to and write an expression for

Explanation:

Step1: Recall the definition of sine in a right - triangle

In a right - triangle, \(\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}\). For \(\angle C\) in the right - triangle with hypotenuse \(a\) and opposite side \(h\), \(\sin(C)=\frac{h}{a}\). For \(\angle A\) in the right - triangle with hypotenuse \(c\) and opposite side \(h\), \(\sin(A)=\frac{h}{c}\).

Step2: Analyze Bret's work

Bret's first step \(\sin(C)=\frac{c}{h}\) and \(\sin(A)=\frac{a}{h}\) is incorrect. The correct formula for sine in a right - triangle is \(\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}\), not \(\frac{\text{hypotenuse}}{\text{opposite}}\).

Step3: Analyze Kamala's work

Kamala starts with \(\sin(C)=\frac{h}{a}\), then \(a\sin(C) = h\) (by cross - multiplying \(a\) on both sides of \(\sin(C)=\frac{h}{a}\)). Also, \(\sin(A)=\frac{h}{c}\), then \(c\sin(A)=h\) (by cross - multiplying \(c\) on both sides of \(\sin(A)=\frac{h}{c}\)). Since \(a\sin(C)=h\) and \(c\sin(A)=h\), we can set \(a\sin(C)=c\sin(A)\), and then divide both sides by \(ac\) to get \(\frac{\sin(C)}{c}=\frac{\sin(A)}{a}\)

Answer:

Kamala