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The angle α is 15°.
We are going to solve the puzzle by using trigonometry. Consider the shaded triangle. We can fill in the third angle and set the square side length to 1 for convenience. The diagonal then has length √2.
We can apply the Law of sines to get sin(180-3α)=sin(3α)=√2sin(2α). We use the triple- and double-angle equalities sin(3α)=3sin(α)-4sin3(α) and sin(2α)=2sin(α)cos(α) to get sin(α)[3-4sin2(α)-2√2cos(α)]=0.
We get from the diagram that α is larger than 0° and smaller than 45°. We can therefore discard the common sine-term. The other factor can be rewritten using sin2(α)+cos2(α)=1 as 4cos2(α)-2√2cos(α)-1=0.
This quadratic equation has two roots: cos(α)=√2(1±√3)/4, corresponding to α=15° and α=105°. The latter can be discarded.