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my problem has two different parts to it. I've made the diagram below to illustrate the context. My questions are:
1) Given the information below, I'd liked to know how to find the angle of EBC
2) Given the information below, I'd also like to know how to find the length of AE
Thanks! My geometry is a bit rusty and I'd really like to figure out how to do this problem.

enter image description here

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3 Answers 3

BE = BC = 8

AB = 2

Right triangle, two known sides.....

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Ah! I didn't realize EB would equal BC. Thanks! –  hammerbrostime Dec 5 '13 at 19:46

EB and BC have equal lengths, so from your data the small triangle ABE has hypotenuse of length 8, small cathetus of length 2 and you can find the length of AE by Phytagora's theorem. Once you know the sides of the small triangle, you can get its angles in a minute. Finally the angle EBC is $\pi$ minus the angle ABE.

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Hints:

$$BC=BE=8\implies AB=2\stackrel{\text{Pythagoras}}\implies AE=\sqrt{8^2-2^2}=2\sqrt{15}\implies$$

$$EC=\sqrt{60+100}=4\sqrt{10}$$

Thus, $\;\Delta BEC\;$ is an isosceles triangle with sides $\;8,8,4\sqrt{10}\;$ and now you can use simple geometry to answer your question (further hint: $\;\angle EBC\;$ is an exterior angle to $\;\Delta ABE\;$ ...)

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Thanks, for the play by play - very clear! –  hammerbrostime Dec 5 '13 at 19:50

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