Z sumy kątów w trójkącie:
|∡ABC| = 180° - 90° - 35° = 55°
Z twierdzenia sinusów:
[tex]\dfrac{|AC|}{\sin|\angle ABC|}=\dfrac{|BC|}{\sin|\angle BAC|}\\\\\\ \dfrac{|AC|}{\sin55^o} =\dfrac2{\sin35^o}\qquad/\cdot\sin55^o\\\\\\ |AC|=\dfrac{2\sin55^o}{\sin35^o}=\dfrac{2\cdot0,8192}{0,5736}\approx2,856[/tex]
Trzeci bok obliczymy z twierdzenia Pitagorasa:
[tex]|AC|^2+|BC|^2=|AB|^2\\\\2{,}856^{\,2}+2^2=|AB|^2\\\\|AB|^2=8,156736+4 \\\\ |AB|^2=12,156736\\\\ |AB|\approx3,487[/tex]
Z jedynki trygonometrycznej mamy:
[tex]\sin^2\alpha+\cos^2\alpha=1\qquad/-\sin^2\alpha\\\\\cos^2\alpha=1-\sin^2\alpha[/tex]
Zatem:
[tex]2\sin^2\alpha-4\cos^2\alpha=2\sin^2\alpha-4\cdot(1-\sin^2\alpha)=2\sin^2\alpha-4+4\sin^2\alpha=\\\\=6\sin^2\alpha-4=6\cdot\left(\dfrac{\sqrt3}2\right)^2-4\, =\, ^3\,{\not}6\cdot\dfrac3{{\not}4_{\,2}}-4\,=\,\dfrac92-\dfrac82\,=\,\dfrac12[/tex]
[tex]\sin\alpha=\dfrac{\sqrt3}2\quad\implies\quad\alpha=60^o\\\\\cos60^o=\dfrac12[/tex]
Zatem: [tex]2\sin^2\alpha-4\cos^2\alpha=2\cdot\left(\dfrac{\sqrt3}2\right)^2-4\cdot\left(\dfrac12\right)^2 =2\cdot\dfrac34-4\cdot\dfrac14=\dfrac32-1=\dfrac12[/tex]
