Ta có :
\(\widehat{A}+\widehat{B}+\widehat{C}=180^o\)
\(\Leftrightarrow\widehat{A}=180^o-\widehat{B}-\widehat{C}\)
\(\Leftrightarrow\widehat{A}=180^o-45^{^{ }o}-30^o=105^o\)
Theo định lý hàm sin ta có :
\(\dfrac{BC}{sinA}=\dfrac{AC}{sinB}\)
\(\Leftrightarrow\left|\overrightarrow{BC}\right|=BC=\dfrac{AC}{sinB}.sinA\left(1\right)\)
\(sinA=sin105^o=sin\left(90^o+15^o\right)=cos15^o\)
\(cos30^o=2cos^215^o-1\)
\(\Leftrightarrow2cos^215^o=cos30^o+1\)
\(\Leftrightarrow cos^215^o=\dfrac{cos30^o+1}{2}\)
\(\Leftrightarrow cos^215^o=\dfrac{\dfrac{\sqrt[]{3}}{2}+1}{2}=\dfrac{\sqrt[]{3}+2}{4}\)
\(\Leftrightarrow cos15^o=\dfrac{\sqrt[]{\sqrt[]{3}+2}}{2}\left(0^o< 15^o< 90^o\right)\)
\(\left(1\right)\Leftrightarrow\left|\overrightarrow{BC}\right|=BC=\dfrac{8a}{\dfrac{\sqrt[]{2}}{2}}.\dfrac{\sqrt[]{\sqrt[]{3}+2}}{2}\)
\(\Leftrightarrow\left|\overrightarrow{BC}\right|=BC=\dfrac{8a\sqrt[]{2}}{2}.\sqrt[]{\sqrt[]{3}+2}\)
\(\Leftrightarrow\left|\overrightarrow{BC}\right|=BC=4a\sqrt[]{\sqrt[]{2}\left(\sqrt[]{3}+2\right)}\)
