Diketahui segitiga ABC dengan A (-3,0)B(0,-6) dan c (3,0)

AB = c = √(0-(-3))²+(-6 – 0)²
= √9 + 36
= √45
= 3√5
BC = a = √(3-0)²+(0-(-6))²
= √9 + 36
= √45
= 3√5
AC = b = √(-3 – 3)²+(0-0)²
= √36 + 0
= 6

cos A = b² + c² – a² / 2bc
= 6² + (3√5)²- (3√5)² / 2 (6) (3√5)
= 36 + 45 – 45 / 36√5
= 36 / 36√5
= 0,447
cos A = 63,4°

Hmm… pertanyaanya apa ya?….