Given
\(\vec A = 3\)
\(\vec B = 4\)
\(\vec C = 5\)
\(\vec A + \vec B = \vec C\)
Squaring both sides
\(|\vec A|^2 + |\vec B|^2 + 2\vec A.\vec B = \vec C .\vec C\)
\(|\vec A|^2 + |\vec B|^2 + 2|\vec A||\vec B| \,cos \theta = |\vec C|^2\)
\((3)^2 + (4)^2 + 2\times 3 \times 4 \, cos \theta = (5)^2\)
\(cos\theta = 0\)
\(\theta = 90° \)