# Thread: Snell's law using triangles,

1. ## Snell's law using triangles,

Hi,

Given two right angle triangles sharing a common side AC.

First triangle angle ACB is $\displaystyle \frac{ \pi }{2}-\theta$
angle ABC is a right angle.

Second triangle angle CAD is $\displaystyle \frac{ \pi }{2}-\alpha$
angle ADC is a right angle.

Why is the following statement true.

$\displaystyle \frac{BC}{cos(\frac{\pi}{2}-\theta)} = \frac{AD}{cos(\frac{\pi}{2}-\alpha)}$

I don't see how this can hold when $\displaystyle \theta \neq \alpha$

I tried using pythagoras, but that doesnt' work.
Perhaps I am missing something about the angle of refraction.

2. We have $\displaystyle \widehat{BAC}=\theta, \ \widehat{ACD}=\alpha$

In triangle ABC use the sine law.

$\displaystyle \frac{BC}{\sin\theta}=\frac{AC}{\sin B}$

But $\displaystyle \sin\theta=\cos\left(\frac{\pi}{2}-\theta\right)$ and $\displaystyle \sin B=1$

Then $\displaystyle \frac{BC}{\cos\left(\frac{\pi}{2}-\theta\right)}=AC$

$\displaystyle \frac{AD}{\cos\left(\frac{\pi}{2}-\alpha\right)}=AC$