Find the angle of a triangle I'v tried to solve this problem but did not get the right result. Triangel PQR is PQ = 5,0 cm, QR = 6,3 cm and RP = 7,4 cm. Calculate angle P. I tried to solve it by using by using the following formula $c^2 = a^2 + b^2 - 2abcosP$. The result I get is 41 degrees but the correct result seems to be 57 degrees. I used the following steps. $cosP = (a^2+b^2-c^2)/(2 \cdot a \cdot b )$ Thanks!

Indeed the solution I get is $\;57.215^\circ\;$ . You have the right formula but maybe you messed between the sides. It should be

$$\cos\alpha=\frac{5^2+7.4^2-6.3^2}{74}$$