Math Help Forum - View Single Post

Risrocks · #5 June 18th, 2008, 03:23 AM

Quote:

Originally Posted by Reckoner

I honestly can't make heads or tails of your work. It isn't clear to me at all what you are doing. Where did $ab = 2\sqrt{10^2 + 10^2}$ come from? And you can't use the law of cosines that way--you need 3 sides of the same triangle.

Anyway, here is how I would do it:

First, I suggest finding the height of the "roof" (that is, the length of the perpendicular between $a$ and $b$ ). Call it $h$ , and we have:

$\tan22^\circ = \frac h{10000+15000}$

$\Rightarrow h = 25000\tan22^\circ\approx10100.66$

Now everything follows from the Pythagorean theorem:

$a^2 = 10000^2 + 25000^2\tan^222^\circ$

$\Rightarrow a = 5000\sqrt{25\tan^222^\circ+4}\approx14213.488$

$c$ is found in much the same way, and $b = a$ since the two inner right triangles are congruent (two sides and an included angle equal).

By the way ... this is how frustrating this is for me (and you having to answer me)! I can't grasp how from:
$a^2 = 10000^2 + 25000^2\tan^222^\circ$
you get the
5000\sqrt etc... etc...
I should just give up right?