tolong bantu dgn cara penyelesaiannya :)

a)
tan A = -1
tan A = tan 135°
maka
A = 135° ± k×180°
k = 0 → A = 135°
k = 1 → A = 315°

jadi A = 135° atau A = 315°

b)
sin (A - 30°) = 1/√2
sin (A - 30°) = sin 45°
maka

1) A - 30° = 45° + k×360°
A = 75° + k×360°
k = 0 → A = 75°
2) A - 30° = 180°-45° + k×360°
A = 165° + k×360°
k = 0 → A = 165°

jadi A = 75° atau A = 165°

c) cos (A - 20°) = -½
cos (A - 20°) = cos 120°
maka

1) A - 20° = 120° + k×360°
A = 140° + k×360°
k = 0 → A = 140°
2) A - 20° = -120° + k×360°
A = -100° + k×360°
k = 1 → 260°

jadi A = 140° atau A = 260°

semoga membantu