Ars Mathematica: Solución 3 problema 17

\overline{B D} = D C

A D = m

A D C

\frac{a}{s e n 15 º} = \frac{m}{s e n 30 º}

\overset{▵}{A B D}

\frac{a}{s e n α} = \frac{m}{s e n (135 - α)}

\frac{s e n 15}{s e n 30} = \frac{s e n α}{s e n (135 - α)} \Rightarrow \frac{s e n α s e n 30}{s e n 15} = s e n (135 - α)

s e n (135 - α) = s e n 135 c o s (- α) + c o s 135 s e n (- α) = \frac{\sqrt{2}}{2} c o s α + \frac{\sqrt{2}}{2} s e n α

c o s α

\frac{t g α}{2 s e n 15} = \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} t g α \Rightarrow t g α = 2 s e n 15 \frac{\sqrt{2}}{2} (1 + t g α)

t g α = \sqrt{2} s e n 15 + \sqrt{2} s e n 15 t g α \Rightarrow t g α - \sqrt{2} s e n 15 t g α = \sqrt{2} s e n 15

t g α (1 - \sqrt{2} s e n 15) = \sqrt{2} s e n 15 \Rightarrow t g α = \frac{\sqrt{2} s e n 15}{1 - \sqrt{2} s e n 15} (2)

s e n \frac{α}{2} = \sqrt{\frac{1 - c o s α}{2}} \Rightarrow s e n 15 = \sqrt{\frac{1 - \frac{\sqrt{3}}{2}}{2}} = \frac{\sqrt{2 - \sqrt{3}}}{2} = \frac{\sqrt{6}}{2} - \frac{\sqrt{2}}{2}

\sqrt{2} (\frac{\sqrt{6}}{2} - \frac{\sqrt{2}}{2}) = \frac{\sqrt{3}}{2} - \frac{1}{2}

\frac{\sqrt{2} s e n 15}{1 - \sqrt{2} s e n 15} = \frac{\sqrt{2} (\frac{\sqrt{6}}{2} - \frac{\sqrt{2}}{2})}{1 - \sqrt{2} (\frac{\sqrt{6}}{2} - \frac{\sqrt{2}}{2})} = \frac{\frac{\sqrt{3}}{2} - \frac{1}{2}}{1 - \frac{\sqrt{3}}{2} - \frac{1}{2}} = \frac{\frac{\sqrt{3}}{2} - \frac{1}{2}}{\frac{2 - \sqrt{3} + 1}{2}} = \frac{\sqrt{3} - 1}{3 - \sqrt{3}} =

= \frac{(\sqrt{3} - 1) (3 + \sqrt{3})}{(3 - \sqrt{3}) (3 + \sqrt{3})} = \frac{3 \sqrt{3} - 3 + 3 - \sqrt{3}}{9 - 3} = \frac{2 \sqrt{3}}{6} = \frac{\sqrt{3}}{3} \Rightarrow α = 30 º

Ars Mathematica