По теореме синусов:
∠C=180°-(60°+45°)=75°
BC/sin60°=AC/sin75°
(√6)/(√3)/2=AC/(√3+1)/2√2
AC=(√6×√3/2)/(√3+1)/2√2=(√6)/1/2
AC=6
ответ:6
По теореме синусов:
∠C=180°-(60°+45°)=75°
BC/sin60°=AC/sin75°
(√6)/(√3)/2=AC/(√3+1)/2√2
AC=(√6×√3/2)/(√3+1)/2√2=(√6)/1/2
AC=6
ответ:6