Răspuns:
1)ABC dreptunghic in A, deci conform Teoremei lui Pitagora BC^2=AB^2+AC^2
BC^2=900+256=1156
BC=32
sin B=AC/BC=16/32=1/2
sin C=AB/BC=30/32=15/16
tg B=AC/AB=16/30=8/15
tg C=AB/AC=30/16=15/8
2) ABC dreptunghic in A si sin B=3/4
sin^2B+cos^2B=1
deci cos^2B=1-9/16=7/16-->cos B=[tex]\sqrt{7} [/tex]/4=sin C(unghiuri complementare)
sin B=cos C=3/4
tg B=sin B/cos B=3*[tex]\sqrt{7} [/tex]/7
tg C=sin C/cos C=[tex]\sqrt{7\\ [/tex]/3
