Quote Originally Posted by shanegoeswapow View Post
Added to the questions i have (August 2006) numbers 29...
In \triangle OSE, we have \angle \frac {E}{2} \simeq \angle ESO = \frac {180 - 40}{2} = 20^{ \circ}. Since \angle EOS = m \widehat {SE} = 140 and \triangle OSE is isoseles.

So \angle E = 40^{ \circ}

Since \triangle ESA is isoseles, \angle S = \angle A = \frac {180 - 40}{2} = 70^{ \circ}


Finally, by the law of Sines:

\frac {SA}{ \sin 40^{ \circ}} = \frac {10}{ \sin 70^{ \circ}}

\Rightarrow SA = \frac {10 \sin 40^{ \circ}}{ \sin 70^{ \circ}} \approx 6.8