cos A = -5/13 , sin B = 4/5 .sudut A tumpul, sudut B lancip, maka sin A cos B + cosA sin B =......? terima kasih

Mapel : Matematika
Kelas : X SMA
Bab : Trigonometri

Pembahasan :
Cos A = -5/13
Sin A = 12/13

Sin B = 4/5
Cos B = 3/5

Maka...
Sin A Cos B + Cos A Sin B
= (12/13)(3/5) + (-5/13)(4/5)
= 36/65 - 20/65
= (36 - 20)/65
= 16/65

cos A = -5/13 = samping/miring
depan = √(13^2 - 5^2) = √(169 - 25) = √144 = 12
sin A = depan/miring = 12/13

sin B = 4/5 = depan/miring
samping = √(5^2 - 4^2) = √(25 - 16) = √9 = 3
cos B = samping/miring = 3/5

sin A cos B + cos A sin B
= 12/13 . 3/5 + (-5/13) . 4/5
= 36/65 - 20/65
= 16/65