1. Problem statement: We have a right triangle $\triangle ABC$ with $\angle A = 90^\circ$, side $AC = 4.8$ cm, and hypotenuse $BC = 7.2$ cm. We need to find the angles $\angle B$ and $\angle C$. 2. Formula and rules: In a right triangle, the sum of the angles is $180^\circ$, and since $\angle A = 90^\circ$, the other two angles satisfy $\angle B + \angle C = 90^\circ$. We can use the sine, cosine, or tangent ratios to find the angles. For example, using sine: $$\sin(\theta) = \frac{\text{opposite side}}{\text{hypotenuse}}$$ 3. Find $\angle B$: Side opposite $\angle B$ is $AC = 4.8$ cm, hypotenuse is $BC = 7.2$ cm. $$\sin(B) = \frac{AC}{BC} = \frac{4.8}{7.2}$$ Simplify the fraction: $$\sin(B) = \frac{\cancel{4.8}}{\cancel{7.2}} = \frac{2}{3}$$ 4. Calculate $\angle B$: $$B = \sin^{-1}\left(\frac{2}{3}\right) \approx 41.81^\circ$$ 5. Find $\angle C$ using the angle sum property: $$C = 90^\circ - B = 90^\circ - 41.81^\circ = 48.19^\circ$$ 6. Final answer: $$\angle B \approx 41.81^\circ, \quad \angle C \approx 48.19^\circ$$