- #1
Kamataat
- 137
- 0
I have to prove that a/\sin\alpha =b/\sin\beta. A triangle has sides "a", "b", "c" and angles \alpha and \beta (opposite of the sides "a" and "b" respectively).
This is what I did:
Draw a line "h" as the height of the triangle on the side "c".
\sin\alpha = h/b. Multiplying by "b" gives b\sin\alpha = h
\sin\beta = h/a. Multiplying by "a" gives a\sin\beta = h
From this we see that b\sin\alpha = a\sin\beta and from this
\frac{a}{\sin\alpha}=\frac{b}{\sin\beta}.
Correct?
- Kamataata
This is what I did:
Draw a line "h" as the height of the triangle on the side "c".
\sin\alpha = h/b. Multiplying by "b" gives b\sin\alpha = h
\sin\beta = h/a. Multiplying by "a" gives a\sin\beta = h
From this we see that b\sin\alpha = a\sin\beta and from this
\frac{a}{\sin\alpha}=\frac{b}{\sin\beta}.
Correct?
- Kamataata
