# Trig proof

1. Mar 23, 2005

### Kamataat

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

2. Mar 23, 2005

### dextercioby

It looks okay to me...

Daniel.

3. Mar 23, 2005

### Kamataat

Thanks for the quick help!