Diketahui a,b, dan y sudut-sudut suatu segitiga. Jika sin a= 12/13 dan cos b= -3/5. Nilai sin y =...

sin a = 12/13 = de/mi
sa = √(13^2 - 12^2) = √(169 - 144) = √25 = 5
cos a = sa/mi = 5/13

cos b = -3/5 = sa/mi ==> b tumpul krn cos b bernilai negatif
de = √(5^2 - 3^2) = √(25 - 9) = √16 = 4
sin b = de/mi = 4/5

a + b + y = 180°
y = 180° - (a + b)
sin y = sin (180° - (a + b))
= sin (a + b)
= sin a cos b + cos a sin b
= 12/13 . -3/5 + 5/13 . 4/5
= -36/65 + 20/65
= -16/65