formula for calculating any angle of an isosceles triangle

Question

This might be an easy question to ask, but for some reason I can not find the right formula to calculate one (does not matter which one) angle (red) of an isosceles triangle (green).

Here is an example:

given:

• triangle ABE
• segment AB has a length of 2cm (c)
• segment BE has a length of 1.12cm (a)
• segment AE has the same length as segment BE (b)

looking for:

• either angle AEB (gamma)
• or angle ABE (beta)
• or angle BEA (alpha)

You can find an image of the triangle here.

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Lets see what I tried so far.

I found a formula which says:

cos. of alpha = ( b ^ 2 + c ^ 2 - a ^ 2 ) / 2 * b * c

So I tried:

              = ( 1.12 ^ 2 + 2 ^ 2 - 1.12 ^ ) / 2 * 1.12 * 2
              = 6,5088                        / 4,48
              = 1,452857143

which is the wrong result.

I found another couple formulas (which I do not remember), but they did not work.

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If anyone knows the right formula, I'd be happy to hear it. :)

Answer 1

draw a line from E to the middle of AB. This is the altitude of an isosceles triangle, so it divides AB in half.

$cos(beta) = 1/1.12$

measure of angle beta = $cos^{-1}(1/1.12) = 26.76°$

Answer 2

HINT: Notice, you should use the cosine formula as follows $$\cos\alpha=\frac{a^2+b^2-c^2}{2ab}=\frac{1.12^2+1.12^2-2^2}{2\cdot 1.12\cdot 1.12}$$