Folosim formula [tex]S=\dfrac{a^2 sinB sinC}{2 sinA}[/tex] si observam ca unghiul A are masura de 180-45-60 =75 grade, iar
sin75=sin(30+45)=sin30cos45+sin45cos30=1/2·√2/2+√2/2·√3/2=√2/4(1+√3)
Deci [tex]S=\dfrac{BC^2\ sinB\ sinC}{2\ sinA}=\dfrac{8\cdot\dfrac{\sqrt2}{2}\cdot \dfrac{\sqrt3}{2}}{2\cdot\dfrac{\sqrt2}{4}(\sqrt3+1)}}=\dfrac{4\sqrt3}{\sqrt3+1}=\dfrac{4\sqrt3(\sqrt3-1)}{2}=[/tex]
[tex]=2(\sqrt3-3)[/tex]
