Question 616924
<font face="Times New Roman" size="+2">


If you have a question that involves an "angel", adjacent or otherwise, please ask the local clergyperson or tribal shaman of your personal choice.  In the event you actually meant "angle", read on:


The short leg of a 30-60-90 right triangle always measures one-half the measure of the hypotenuse.  C.f. *[tex \LARGE \sin(30^\circ)\ =\ \frac{1}{2}].


So if you have a right triangle where one leg measures 6, another leg measures *[tex \LARGE \frac{c}{2}] and the hypotenuse measures *[tex \LARGE c], then, according to Mr. Pythagoras:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ c^2\ =\ 6^2\ +\ \left(\frac{c}{2}\right)^2]


Solve for *[tex \LARGE c].


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
<div style="text-align:center"><a href="http://outcampaign.org/" target="_blank"><img src="http://cdn.cloudfiles.mosso.com/c116811/scarlet_A.png" border="0" alt="The Out Campaign: Scarlet Letter of Atheism" width="143" height="122" /></a></div>
</font>