Question

va rog ajutati-ma ca nu știu cum sa rezolv tema​

Answer 1

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=$\sqrt{7}$/4=sin C(unghiuri complementare)

sin B=cos C=3/4

tg B=sin B/cos B=3*$\sqrt{7}$/7

tg C=sin C/cos C=$\sqrt{7$/3