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 July 6th, 2014, 02:55 AM #1 Senior Member   Joined: Nov 2013 Posts: 434 Thanks: 8 the figure
 July 6th, 2014, 03:16 AM #2 Senior Member     Joined: Feb 2010 Posts: 706 Thanks: 140 $\displaystyle \dfrac{20}{\sin C}=\dfrac{10}{\sin 25}$ gives you $\displaystyle C=\sin^{-1}(2 \sin 25)$ and $\displaystyle ADB=180-C$. Saying $\displaystyle m\angle ADB > m \angle C$ is useless since it is always true.

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