Step-by-step explanation:
In ∆ABC, m∠B = 90°
Therefore ∆ABC is a right angled triangle with sides AB = 4 & BC = 3. AC is the hypotenuse.
Hence, by Pythagoras theorem:
[tex]AC^2 = AB^2 +BC^2 \\ \\ \therefore \: AC = \sqrt{AB^2 +BC^2 } \\ = \sqrt{ {4}^{2} + {3}^{2} } \\ = \sqrt{16 + 9} \\ = \sqrt{25} \\ = 5 \\ \\ \huge \red{ \boxed{ \therefore \: AC = 5 \: units}}[/tex]