tlng dibantu, tentukan nilai perbandingan trigonometri dari sudut A dan B, nyatakan jawaban dalam bentuk paling sederhana. dengan cara penyelesaiannya makasih s

1) AB^2 = AC^2 + BC^2
AB^2 = (2V3)^2 + 2^2
AB^2 = 12 + 4
AB^2 = 16
AB = V16
AB = 4
sin A = BC/AB = 2/4 =/1/2
cos A = AC/AB = 2V3/4 = (1/2)V3
tan A = BC/AC = 2/2V3 = 1/V3 = (1/3)V3
sin B = AC/AB = 2V3/4 = (1/2)V3
cos B = BC/AB = 2/4 = 1/2
tan B = AC/BC = 2V3/2 = V3
2) AC^2 = AB^2 - BC^2
AC^2 = 7^2 - 5^2
AC^2 = 49 - 25
AC^2 = 24
AC = V24 = 2V6
cos A = AC/AB = 2V6/7 = (2/7)V6
sin A = BC/AB = 5/7
tan A = BC/AC = 5/2V6 = (5/12)V6
sin B = AC/AB = 2V6/7 = (2/7)V6
cos B = BC/AB = 5/7
tan B = AC/BC = 2V6/5 = (2/5)V6