Question

bret
\\( \sin ( c ) = \frac { c } { h } \\)
\\( \sin ( a ) = \frac { a } { h } \\)
\\( h = \frac { a } { \sin ( a ) } \\)
\\( h = \frac { c } { \sin ( c ) } \\)
\\( \frac { a } { \sin ( a ) } = \frac { c } { \sin ( c ) } \\)
\\( \frac { \sin ( c ) } { c } = \frac { \sin ( a ) } { a } \\)
kamala
\\( \sin ( c ) = \frac { h } { a } \\)
\\( \sin ( a ) = \frac { h } { c } \\)
\\( a \sin ( c ) = h \\)
\\( c \sin ( a ) = h \\)
\\( a \sin ( c ) = c \sin ( a ) \\)
draw in a new perpendicular line
use the pythagorean theorem
write an expression for the cosine of a
the proof was correctly comple the next step in the proof is to and write an expression for

Explanation:

Step1: Analyze Bret's and Kamala's work

Bret and Kamala have derived expressions for \(h\) in terms of the sides and sines of angles in the triangle. Bret uses \(\sin(C)=\frac{c}{h}\) (incorrect, should be \(\sin(C)=\frac{h}{a}\) in the right - triangle with side \(a\) as the hypotenuse) and \(\sin(A)=\frac{a}{h}\) (incorrect, should be \(\sin(A)=\frac{h}{c}\) in the right - triangle with side \(c\) as the hypotenuse). Kamala correctly uses the sine definition in right - triangles: \(\sin(C)=\frac{h}{a}\) (in the right - triangle with hypotenuse \(a\)) and \(\sin(A)=\frac{h}{c}\) (in the right - triangle with hypotenuse \(c\)). Then she gets \(a\sin(C)=h\) and \(c\sin(A)=h\), so \(a\sin(C)=c\sin(A)\).

Step2: Generalize the law of sines

To complete the law of sines proof for \(\frac{a}{\sin(A)}=\frac{b}{\sin(B)}=\frac{c}{\sin(C)}\), we can draw a new perpendicular line from another vertex (say, from \(B\) to \(AC\)) and repeat the process.

Answer:

draw in a new perpendicular line