sinA = sin(π-(B+C))= sin(B+C)=(sinBcosC+cosBsinC)=√2(2√5/5+√5/5)/2 = 3√ 10/ 10。
(2) By sine theorem: BC=ACsinA/sinB=6. CD=3。
By cosine theorem: AD=√(AC? +CD？ -2*AC*CD*cosC)=√5。
(The above "√" indicates the root sign)