Đáp án:
$\begin{array}{l}
a)\widehat {BOC} = 2.\widehat {BAC} = {2.60^0} = {120^0}\\
\widehat {COA} = 2.\widehat {CBA} = {2.70^0} = {140^0}\\
\Rightarrow \widehat {AOB} = {360^0} - {120^0} - {140^0} = {100^0}\\
b)\widehat C = {180^0} - {60^0} - {70^0} = {50^0}\\
\Rightarrow \widehat C < \widehat A < \widehat B\left( {{{50}^0} < {{60}^0} < {{70}^0}} \right)\\
\Rightarrow cung\,AB < cung\,BC < cung\,CA\\
c)OH \bot BC\\
\Rightarrow \widehat {BOH} = \widehat {COH} = \frac{1}{2}\widehat {BOC} = {60^0}\\
Trong:\Delta BOH \bot H\\
\Rightarrow \widehat {OBH} = {30^0}\\
\Rightarrow OH = \frac{1}{2}BO = \frac{1}{2}R\\
\Rightarrow BH = \sqrt {O{B^2} - O{H^2}} = \sqrt {{R^2} - \frac{1}{4}{R^2}} = \frac{{\sqrt 3 }}{2}R\\
\Rightarrow BC = 2BH = \sqrt 3 R
\end{array}$
