Răspuns :
Construim AH perpendicular pe BC.
In triunghiul AHB dreprtunghic in AHB avem:
sinB=AH/AB =>√3 /2=AH/4 => AH=2√3(cm)
cosB=BH/AB => 1/2=BH/4 => BH=2(cm) =>CH=4(cm)
In triunghiul AHC dreptunghic in AHC avem:
conform Th Pitagora: AC²=AH²+HC² => AC²=12+16 => AC=2√7(cm)
sinC=AH/AC=(2√3)/(2√7) => sinC=√21/7
In triunghiul AHB dreprtunghic in AHB avem:
sinB=AH/AB =>√3 /2=AH/4 => AH=2√3(cm)
cosB=BH/AB => 1/2=BH/4 => BH=2(cm) =>CH=4(cm)
In triunghiul AHC dreptunghic in AHC avem:
conform Th Pitagora: AC²=AH²+HC² => AC²=12+16 => AC=2√7(cm)
sinC=AH/AC=(2√3)/(2√7) => sinC=√21/7
