1. **State the problem:** Given a triangle with side $a=37$, side $c=40$, and angle $\angle A=35^\circ$, find the possible values of angles $\angle B_1$, $\angle B_2$, $\angle C_1$, $\angle C_2$, and sides $b_1$, $b_2$ using the Law of Sines. 2. **Formula and rules:** The Law of Sines states: $$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$$ Since two sides and one angle are given, we can find angle $B$ using: $$\sin B = \frac{b}{a} \sin A$$ But here, $b$ is unknown, so we use the Law of Sines between $a$ and $c$: $$\frac{a}{\sin A} = \frac{c}{\sin C}$$ From this, we find $\sin C$ and then $\angle C$. 3. **Calculate $\sin C$ and $\angle C$:** $$\sin C = \frac{c}{a} \sin A = \frac{40}{37} \sin 35^\circ$$ Calculate $\sin 35^\circ \approx 0.574$: $$\sin C = \frac{40}{37} \times 0.574 = 1.081 \times 0.574 = 0.620$$ 4. **Find $\angle C_1$ and $\angle C_2$:** Since $\sin C = 0.620$, possible angles are: $$\angle C_1 = \arcsin(0.620) \approx 38.3^\circ$$ $$\angle C_2 = 180^\circ - 38.3^\circ = 141.7^\circ$$ 5. **Find $\angle B_1$ and $\angle B_2$:** Sum of angles in triangle is $180^\circ$: $$\angle B_1 = 180^\circ - \angle A - \angle C_1 = 180^\circ - 35^\circ - 38.3^\circ = 106.7^\circ$$ $$\angle B_2 = 180^\circ - 35^\circ - 141.7^\circ = 3.3^\circ$$ 6. **Calculate sides $b_1$ and $b_2$ using Law of Sines:** $$b_1 = \frac{a \sin B_1}{\sin A} = \frac{37 \sin 106.7^\circ}{\sin 35^\circ}$$ Calculate $\sin 106.7^\circ \approx 0.972$, $\sin 35^\circ \approx 0.574$: $$b_1 = \frac{37 \times 0.972}{0.574} = \frac{35.96}{0.574} \approx 62.6$$ Similarly for $b_2$: $$b_2 = \frac{37 \sin 3.3^\circ}{\sin 35^\circ}$$ Calculate $\sin 3.3^\circ \approx 0.058$: $$b_2 = \frac{37 \times 0.058}{0.574} = \frac{2.15}{0.574} \approx 3.75$$ **Final answers:** - $\angle B_1 \approx 106.7^\circ$ - $\angle B_2 \approx 3.3^\circ$ - $\angle C_1 \approx 38.3^\circ$ - $\angle C_2 \approx 141.7^\circ$ - $b_1 \approx 62.6$ - $b_2 \approx 3.75$