Jika: sin A= 3/5 dan tan B=1/7 dgn A dan B sudut lancip, tunjukkan bahwa A+B=45° , plis mohon banget bantuannyaa
whongaliem
Sin A = 3/5 ; tan B = 1/7 sin² A + cos² A = 1 (3/5)² + cos² A = 1 9/25 + cos² A = 1 cos² A = 1 - 9/25 cos² A = 16/25 cos A = √(16/25) cos A = 4/5 tan A = sin A / cos A = (3/5) / (4/5) = 3/5 x 5/4 = 3/4
tan (A + B) = (tan A + tan B) / (1 - tan A .tan B) = (3/4 + 1/7) / (1 - 3/4 . 1/7) = (25/28) / (1 - 3/28) = (25/28) / (25/28) tan (A + B) = 1 tan (A + B) = tan 45° ∴ A + B = 45° .... terbukti ..
sin² A + cos² A = 1
(3/5)² + cos² A = 1
9/25 + cos² A = 1
cos² A = 1 - 9/25
cos² A = 16/25
cos A = √(16/25)
cos A = 4/5
tan A = sin A / cos A
= (3/5) / (4/5)
= 3/5 x 5/4
= 3/4
tan (A + B) = (tan A + tan B) / (1 - tan A .tan B)
= (3/4 + 1/7) / (1 - 3/4 . 1/7)
= (25/28) / (1 - 3/28)
= (25/28) / (25/28)
tan (A + B) = 1
tan (A + B) = tan 45°
∴ A + B = 45° .... terbukti ..