a)ΔABC dreptunghic⇒m(ACB)=90⁰
m(BAC)=2*m(CBA)
m(ACB)+m(BAC)+m(CBA)=180⁰
90⁰+2*m(CBA)+m(CBA)=180⁰
2*m(CBA)+m(CBA0=180⁰-90⁰=90⁰
3*m(CBA)=90⁰
m(CBA)=90⁰/3=30⁰
m(BAC)=2*30⁰=60⁰
b)ΔCBD(m(CDB)=90⁰):cos (DBC)=DB/CB
cos30⁰=12/CB
√3/2=12/CB⇒CB=12*2/√3=8√3cm
ΔABC(m(ACB)=90°):sin (ACB)=CB/AB
sin60° =8√3/AB
√3/2=8√3/AB⇒AB=8√3*2/√3=16 cm