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I know the solution that is 30 degrees here but I need to know the method so that I can extend it to a general solution. A method with linear equations will be very helpful.

ABCD is a square. Isosceles triangle ABE has base angles measuring 15 degrees. We need to find the angle ECB namely alpha.

Here is the image of the diagram.We need to find the unknown angle alpha.enter image description here

Thomas Andrews
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L Lawliet
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2 Answers2

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Construct the mid line PQ as shown.

enter image description here

Let the square ABCD be $2 \times 2$; and E’ be a point on PQ such that triangle BE’C is equilateral.

Then, $\triangle PE’C$ is an $1 – 2 – \sqrt 3$ triangle with ($\gamma = 60^0$ and $\beta = 30^0$ PC = 1 and CE’ = 2) as shown.

Therefore, $\triangle CE’D$ is isosceles with $\angle CDE’ = 75^0$

Hence, $\angle ADE’ = 15^0$ indicating E’ is actually E.

Mick
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  • This is a good argument, but I don't think you need the second paragraph at all since the lengths $CE'$ and $CD$ give you that it's isoceles. Also, your diagram has lots of extraneous symbols. – Suzu Hirose Aug 09 '22 at 05:05
  • @Suzu Hirose Had you worked with Geogebra or Desmos, you should know that this "lot of extraneous symbols" are due to the default mode which displays the names for everything ! – Jean Marie Aug 09 '22 at 07:20
  • @SuzuHirose In fact, I have some of the symbols hidden already. – Mick Aug 09 '22 at 08:12
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Direct proof using Pythagoras Theorem:

  1. Connect the midpoint ($M$) of the square base to the point $E$.
  2. Let the distance between point $E$ and the triangle base have length $x$.
  3. Now, it should be clear that $$\tan(15) = \frac{x}{1/2} = 2x.$$ and $$\tan(\alpha) = \frac{1/2}{1-x} = \frac{1}{2-2x} = \frac{1}{2-\tan(15)}.$$

Now, to compute $\tan(15)$, we utilize the identity for tangent. $$\tan(\alpha) = 2\frac{\tan(\alpha/2)}{1-\tan(\alpha/2)^2}$$ so that $$\frac{1}{\sqrt{3}} = \tan(30) = 2\frac{\tan(15)}{1-\tan(15)^2}.$$ Letting $x = \tan(15)$ and rearranging, we find the quadratic $$-\frac{1}{2\sqrt{3}}x^2-x+\frac{1}{2\sqrt{3}} = 0.$$ Solving gives, $$x = 2-\sqrt{3}.$$ So, $$\tan(\alpha) = \frac{1}{2-(2-\sqrt{3})} = \frac{1}{\sqrt{3}}.$$ Therefore, $\alpha = 30$.

Doug
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