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

So $\displaystyle \angle E = 40^{ \circ}$

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


Finally, by the law of Sines:

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

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