Extended Law of Sines

If $\triangle{ABC}$ is a triangle with circumradius R, then

$\frac{BC}{sin(\angle{BAC})}$ = $\frac{AC}{sin(\angle{ABC})}$ = $\frac{AB}{sin(\angle{ACB})}$ =2R

2. https://www.geogebra.org/m/mHCYrmXy

3 Comments

Armando

October 27, 2016 at 10:28 pm

Hey Evelin,
Your link isn’t quite working for me, do you mind posting another?

Thank You

Mei Zhu
October 29, 2016 at 2:49 pm

I found the same problem.

Reply

Evelin (Post author)

October 30, 2016 at 4:51 pm

https://ggbm.at/mHCYrmXy

Here’s the link to my dynamic worksheet. Sorry for that guys.

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