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MCIII Geometry 5.7

 
 
ShiJames的头像
MCIII Geometry 5.7
ShiJames - 2020年11月15日 Sunday 12:50
 

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.

 
ShiJames的头像
Re: MCIII Geometry 5.7
ShiJames - 2020年11月15日 Sunday 13:18
 

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

WangDr. Kevin的头像
Re: MCIII Geometry 5.7
WangDr. Kevin - 2020年11月15日 Sunday 14:11
 

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