+73 I'm stuck on this and don't know where to start. How do you do this step by step?
+37146 Well, obviously, she shoud aim at point E..... haha jk
Distance D = (12+9+9) - 27 = 3
The two angles at E are equal (angle of incidence = angle of reflection, generally)
the large and the smaller triangle are similar triangles.....
Does all of that give ya a clue? 
+73 Ummm... I'm not quite sure. What I'm supposed to do from there? Maybe explain how to get where to aim at the wall?
+37146 Similar triangles....
3 is to 12 as CE is to EB AND CE+ EB =20
3/12 = CE/EB
1/4 = CE/ EB Want me to finish it or can you try from here? Do you remember how to solve ratios?
+73 Yeah, can you finish it from there? This is supposed to be the last question on my work. I do know about the ratios but similar tringles always get me. Thanks!
+37146 I have reduced the similar triangles it to a simple ratio problem....you should be able to solve it:
1/4 = CE/EB and CE + EB = 20
so out of FIVE parts (1+4) CE is ONE of them and EB is FOUR of them
20 divided by 5 parts = 4 per part
so CE = 4 EB = 16 CE/EB = 4/16 = 1/4 DO you see how to do ratios nw?
+129852 Note that angles EBA and ECD are equal right angles
And angle BEA = angle CED
So triangle EBA is similar to triangle ECD
So
BE / BA = CE / DC
BA = 12
Let EB = 20 - x
Let CE = x
DC = (12 + 9 + 9) - 27 = 3
So we have that
[20 - x] / 12 = x / 3 cross-multiply
3[20 - x] = 12x simplify
60 - 3x = 12x rearrange as
60 = 15x divide both sides by 15
x = 4
She should aim 4 ft to the right of "C"
EDIT TO CORRECT A PREVIOUS MISTAKE.....THANKS TO EP FOR CATCHING THIS !!!!
+73 Thank you very much CPhill this is just what I needed to complete my work. Very understandable too and I saw your helpful correction to figure out the answer to the problem.
