Xét ΔOAB có OA=OB=AB(=R)
nên ΔOAB đều
=>\(\widehat{AOB}=60^0\)
=>\(\widehat{ACB}=\dfrac{\widehat{AOB}}{2}=30^0\)
Xét ΔOAC có \(cosAOC=\dfrac{OA^2+OC^2-AC^2}{2\cdot OA\cdot OC}\)
\(=\dfrac{R^2+R^2-\left(R\sqrt{3}\right)^2}{2\cdot R\cdot R}=\dfrac{2R^2-3R^2}{2R^2}=-\dfrac{1}{2}\)
nên \(\widehat{AOC}=120^0\)
Vì \(\widehat{ABC}>90^0\)
nên \(\widehat{ABC}=180^0-\dfrac{120^0}{2}=120^0\)
Xét ΔABC có
\(cosABC=\dfrac{BA^2+BC^2-AC^2}{2\cdot BA\cdot BC}\)
=>\(\dfrac{R^2+BC^2-\left(R\sqrt{3}\right)^2}{2\cdot R\cdot BC}=cos120=-\dfrac{1}{2}\)
=>\(\dfrac{BC^2-2R^2}{2RBC}=-\dfrac{1}{2}\)
=>\(2BC^2-4R^2=-4RBC\)
=>\(BC^2-2R^2=-2RBC\)
=>\(BC^2+2\cdot BC\cdot R-2R^2=0\)
=>\(BC^2+2\cdot BC\cdot R+R^2=3R^2\)
=>\(\left(BC+R\right)^2=\left(R\sqrt{3}\right)^2\)
=>\(BC+R=R\sqrt{3}\)
=>\(BC=R\sqrt{3}-R\)
\(\dfrac{R}{BC}=\dfrac{R}{R\sqrt{3}-R}=\dfrac{1}{\sqrt{3}-1}=\dfrac{\sqrt{3}+1}{2}\)
