Tentukan unsur unsur segitiga abc jika diketahui hal berikut ini: sisi a=12,sisi b=5, sudutB=24°

A=12
b=5
sudutB=24°
jawab :

a/sinA = b/sinB
12/sinA=5/sin24°
12/sinA=5/0,41
sinA = 12 × 0,41/5
sinA = 0,984
sudut A = 80

sudut total segitiga = 180°
sudut A + B + C = 180°
80 + 24 + C = 180
C = 180 – 104
C = 76

c/sinC = b/SinB
c/sin76 = 5/sin24
c/0,97 = 5/0,41
c = 5/0,41 × 0,97
c = 11,8