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# I'm stuck on this and don't know where to start.

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I'm stuck on this and don't know where to start. How do you do this step by step? Nov 12, 2019

#1
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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? Nov 12, 2019
edited by ElectricPavlov  Nov 12, 2019
#2
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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?

GAMEMASTERX40  Nov 13, 2019
#3
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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?

Nov 13, 2019
edited by ElectricPavlov  Nov 13, 2019
#4
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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!

Nov 13, 2019
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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?

ElectricPavlov  Nov 13, 2019
#6
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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    !!!!   Nov 13, 2019
edited by CPhill  Nov 13, 2019
#7
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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.

GAMEMASTERX40  Nov 14, 2019