<p>Hey guys,</p>
<p>I was wondering if there was a quick and easy way to do this problem, instead of substituting (180-n) in for the two unmarked angles, and working from there.</p>
<p>Is the problem to hard, or to easy...?</p>
<p>How long really is "working from there"?
For one triangle it's 180 - (180-n) = n.
Double it for two triangles together and you get the answer.</p>
<p>A tad faster would be to know ahead of time the Triangles "Key Fact H2" - on p.66 of Barron's "How to prep for the SAT II Level IC", Ira Wolf:
an exterior angle is equal to the sum of the opposite interior angles.
You get n+n=2n right away.</p>
<p>One other approach, not necessarily faster, but works as a back-up on some math and many physics questions, is to "do the extremes".
Assume that one of the variables is getting "very-very" big (or small), and observe what happens to the other variables, or a function, or a shape, etc. A nice example is a post by this forum good Samaritan optimizerdad
<a href="http://talk.collegeconfidential.com/showthread.php?p=1224915&highlight=large#post1224915%5B/url%5D">http://talk.collegeconfidential.com/showthread.php?p=1224915&highlight=large#post1224915</a>.</p>
<p>In your (BB) question you can safely assume that the triangles are isosceles, and n is almost 180 deg. (the result should be the same for any configuration; this is another SAT trick - choosing any "convenient" shapes and numbers if they are not defined/restricted).
Each marked angle becomes equal almost 90 deg., and their total almost 360 deg.
Only answer B. (2n) approximates this number for n~180 deg.</p>
<p>This solution is probably longer, but if you are in a crunch and don't see any other way, it could be a life saver. Well, not really, just a point earner.</p>
<p>Hope this does not unhelp. I mean, confuse.</p>
<p>I know that this was from a really long time ago...but I'm confused because if the exterior angle=sum of oposite interior angles, that would only take care of two of the interior angles. so wouldn't the answer be 4n?</p>
<p>the fastest solution is draw in two lines on top and on bottom making it a quadrilateral, which has 360 degrees total. 180-n gives you the total degrees of the top triangle. theres two of those, so do 360-2(180-n) to get the degree measure of the 4 angles which is 2n.</p>
<p>^
You meant 180-n gives the sum of the two top angles in the top triangle, and 180-n gives the sum of the two bottom angles in the bottom triangle, right?</p>
<p>Anyway, I still stand by the exterior angle approach as the fastests one (see post #3).</p>