Pada segitiga PQR panjang sisi QR=3 akar 6,PR=6 akar 2,PQ=3 akar 2 berapa besar sudut p ?

aturan cosinus
cos P = (PQ² + PR² - QR²) / (2 x PQ x PR)
cos P = ((3√2)² + (6√2)² - (3√6)²) / (2 x 3√2 x 6√2)
cos P = (18 + 72 - 54)/72
cos P = 36/72
cos P = 1/2
sudut P = 60°

QR² = PQ² + PR² - 2(PQ)(PR) cos P
(3akar6)² = (3akar2)² + (6akar2)² - 2(3akar2)(6akar2) cos P
54 = 18 + 72 - 72 cos P
72 cos P = 36
cos P = 36/72
cos P = 1/2
<P = 60°