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I have such a nice way for ##\tan(15°)=2-\sqrt{3}## with the cosine theorem and all just look up the tangent.My solution for Problem 3

γ=45-α/2

β=135-α/2

According to the Law of sines:

##\overline{AD} = \frac {\overline{CD} \sin(45-α/2)}{\sin(α/2)}## 1

##\overline{AD}= \frac {\overline{BD}\sin(135-α/2)}{\sin(α/2)}## 2

multiplying eqs1 and eqs2 and using that ##\overline{AD}^2=\overline{BD} \overline{CD}##,

##\sin(45°-α/2)\sin(135°-α/2)=\sin^2(α/2)##

The solution is α/2=30°: α=60, β=105°, γ=15°.

Dividing eq 1 with eq 2

##\frac{\overline{BD}}{\overline{CD}} = \frac {\sin(45°-α/2)}{\sin(135°-α/2)} =\frac{\sin(15°) }{\sin(105°)} = \tan(15°)##

Using the half angle formula :##\tan^2(x/2) =\frac {1-\cos(x)}{1+\cos(x)}## . With x=30°, ##\tan(15°)=2-\sqrt{3}##.

View attachment 228447

Correct, of course.