Question

help.
1) In Triunghiul ABC

Aria =9
AB =3
BC=12
sin B=?


2) in triunghiul ABC
cos A=?
AB=3
BC=2
AC=4

Answer 1

[tex]a)A= \frac{1}{2}AB*BC*sinB \\ 9= \frac{1}{2}*3*12*sinB \\ 18=36sinB \\ sinB= \frac{18}{36} \\ sinB= \frac{1}{2} \\ b)BC^2=AB^2+AC^2-2AB*AC*cosA \\ 4=9+16-2*3*4*cosA \\ 4-9-16=-24cosA \\ cosA= \frac{21}{24} \\ cosA= \frac{7}{8} [/tex]