## Online Course Discussion Forum

### MCIII Geometry 5.7

MCIII Geometry 5.7

For this question, I applied trig Ceva to get $\frac{\sin{80^\circ}}{\sin{20^\circ}} \cdot \frac{\sin{10^\circ}}{\sin{30^\circ}} \cdot \frac{\sin{\angle CBP}}{\sin{\angle ABP}} = 1$. However, I am not sure how to simplify all of the sines. I have tried double angle, but I'm not sure if that helps. Could I have a hint? Thanks.

Re: MCIII Geometry 5.7

Nevermind, I think I figured it out. I didn't realize $\sin{80^\circ} = \cos{10^\circ}$.

Re: MCIII Geometry 5.7

Sounds good :) this is one of the easier ones.