Question

which student correctly started the proof, and what should that student do next to complete the proof? student 1: (\frac{sin(a)}{a}=\frac{sin(b)}{b}), (bsin(a)=asin(b)), (asin(b) = h), (bsin(a)=h), (sin(b)=\frac{h}{a}), (sin(a)=\frac{h}{b}) student 2: (\frac{sin(a)}{b}=\frac{sin(b)}{a}), (asin(a)=bsin(b)), (bsin(b)=h), (asin(a)=h), (sin(b)=\frac{h}{b}), (sin(a)=\frac{h}{a})

Explanation:

Step1: Recall the law - of - sines concept

The law of sines for a triangle with sides \(a\), \(b\), \(c\) and opposite angles \(A\), \(B\), \(C\) is \(\frac{a}{\sin(A)}=\frac{b}{\sin(B)}=\frac{c}{\sin(C)}\). The key idea in the proof using the height \(h\) of the triangle is that in a non - right triangle, if we consider the height \(h\) from one vertex to the opposite side, we have \(h = a\sin(B)\) and \(h = b\sin(A)\).

Step2: Analyze student work

Both Student 1 and Student 2 start correctly by establishing \(a\sin(B)=b\sin(A)\) from the fact that the height \(h\) of the triangle can be expressed as \(h = a\sin(B)\) and \(h = b\sin(A)\). From \(a\sin(B)=b\sin(A)\), we can divide both sides by \(\sin(A)\sin(B)\) (assuming \(\sin(A)
eq0\) and \(\sin(B)
eq0\)) to get \(\frac{a}{\sin(A)}=\frac{b}{\sin(B)}\).

Answer:

Both students started the proof correctly. To complete the proof, they should divide both sides of the equation \(a\sin(B)=b\sin(A)\) by \(\sin(A)\sin(B)\) (assuming \(\sin(A)
eq0\) and \(\sin(B)
eq0\)) to obtain \(\frac{a}{\sin(A)}=\frac{b}{\sin(B)}\). Then, they can repeat the process by considering the height from another vertex to show \(\frac{b}{\sin(B)}=\frac{c}{\sin(C)}\) and thus prove the law of sines \(\frac{a}{\sin(A)}=\frac{b}{\sin(B)}=\frac{c}{\sin(C)}\).