diberikan segitiga ABC lancip dengan tan a = 12/5 dan sin b = 4/5 nilai cos (A-B) adalah...

tanA = 12/5 → y = 12 dan x = 5
r = √(y² + x²) = √(12² + 5²) = 13
sinA = y/r = 12/13
cosA = x/r = 5/13
sinB = 4/5 → r = √(5² - 4²) = 3
cosB = x/r = 3/5
cos(A - B)= cosA cosB + sinA sinB
= (5/13)(3/5) + (12/13)(4/5)
= 63/65