Question 1117417
<font face="Times New Roman" size="+2">
*[illustration HikeWithTurnCrop.jpg]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{\sin A}{10}\ =\ \frac{\sin C}{12}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sin C\ =\ \frac{12\sin A}{10}]


This is an ambiguous case, but since angle C is clearly obtuse from the scale diagram, we can assert:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ m\angle C\ =\ 180^\circ\ -\ \arcsin\(\frac{12\sin A}{10}\)]


Turn angle is then:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 180^\circ\ -\ m\angle C]


You can do the calculator work.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
<img src="http://c0rk.blogs.com/gr0undzer0/darwin-fish.jpg">
*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  

</font>