1. **State the problem:** Given triangle ABC with sides $b=24$, $c=68$, and angle $C=61^\circ$, solve for angle $B$, angle $A$, and side $a$ using the Law of Sines. 2. **Formula used:** The Law of Sines states: $$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$$ 3. **Find angle $B$:** Using the Law of Sines, $$\frac{b}{\sin B} = \frac{c}{\sin C} \implies \sin B = \frac{b \sin C}{c}$$ Substitute values: $$\sin B = \frac{24 \times \sin 61^\circ}{68}$$ Calculate $\sin 61^\circ \approx 0.8746$: $$\sin B = \frac{24 \times 0.8746}{68} = \frac{20.9904}{68} \approx 0.3087$$ 4. **Calculate angle $B$:** $$B = \sin^{-1}(0.3087) \approx 18.0^\circ$$ 5. **Find angle $A$:** Sum of angles in a triangle is $180^\circ$: $$A = 180^\circ - B - C = 180^\circ - 18.0^\circ - 61^\circ = 101.0^\circ$$ 6. **Find side $a$:** Using Law of Sines again, $$\frac{a}{\sin A} = \frac{c}{\sin C} \implies a = \frac{c \sin A}{\sin C}$$ Calculate $\sin 101.0^\circ \approx 0.9816$: $$a = \frac{68 \times 0.9816}{0.8746} = \frac{66.7488}{0.8746} \approx 76.3$$ **Final answers:** $$B = 18.0^\circ, \quad A = 101.0^\circ, \quad a = 76.3$$