1. **State the problem:** We have a triangle ABC with sides $x-2$, $x$, and $x+2$. The largest angle is $120^\circ$. We need to find $x$, verify the area, and find $\sin A + \sin B + \sin C$. 2. **Find $x$ using the Law of Cosines:** The largest angle is $120^\circ$, so it is opposite the longest side $x+2$. The Law of Cosines states: $$ (x+2)^2 = (x-2)^2 + x^2 - 2(x-2)(x)\cos 120^\circ $$ Recall $\cos 120^\circ = -\frac{1}{2}$. 3. **Substitute and simplify:** $$ (x+2)^2 = (x-2)^2 + x^2 - 2(x-2)(x)\left(-\frac{1}{2}\right) $$ $$ (x+2)^2 = (x-2)^2 + x^2 + (x-2)(x) $$ 4. **Expand each term:** $$ (x+2)^2 = x^2 + 4x + 4 $$ $$ (x-2)^2 = x^2 - 4x + 4 $$ $$ (x-2)(x) = x^2 - 2x $$ 5. **Rewrite the equation:** $$ x^2 + 4x + 4 = (x^2 - 4x + 4) + x^2 + (x^2 - 2x) $$ 6. **Combine like terms on the right:** $$ x^2 + 4x + 4 = x^2 - 4x + 4 + x^2 + x^2 - 2x $$ $$ x^2 + 4x + 4 = 3x^2 - 6x + 4 $$ 7. **Bring all terms to one side:** $$ 0 = 3x^2 - 6x + 4 - x^2 - 4x - 4 $$ $$ 0 = 2x^2 - 10x $$ 8. **Factor out common terms:** $$ 0 = 2x(x - 5) $$ 9. **Solve for $x$:** $$ x = 0 \quad \text{or} \quad x = 5 $$ Since side lengths must be positive and $x-2$ must be positive, $x=5$. --- 10. **Show area is $\frac{15\sqrt{3}}{4}$:** Use formula for area with two sides and included angle: $$ \text{Area} = \frac{1}{2}ab\sin C $$ Here, sides adjacent to angle $120^\circ$ are $x-2=3$ and $x=5$. 11. **Calculate area:** $$ \text{Area} = \frac{1}{2} \times 3 \times 5 \times \sin 120^\circ = \frac{15}{2} \times \frac{\sqrt{3}}{2} = \frac{15\sqrt{3}}{4} $$ --- 12. **Find $\sin A + \sin B + \sin C$:** Use Law of Sines: $$ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R $$ Longest side $c = 7$ opposite $C=120^\circ$. 13. **Find circumradius $R$:** $$ 2R = \frac{c}{\sin C} = \frac{7}{\sin 120^\circ} = \frac{7}{\frac{\sqrt{3}}{2}} = \frac{14}{\sqrt{3}} $$ 14. **Find $\sin A$ and $\sin B$:** $$ \sin A = \frac{a}{2R} = \frac{3}{\frac{14}{\sqrt{3}}} = \frac{3\sqrt{3}}{14} $$ $$ \sin B = \frac{b}{2R} = \frac{5}{\frac{14}{\sqrt{3}}} = \frac{5\sqrt{3}}{14} $$ 15. **Sum the sines:** $$ \sin A + \sin B + \sin C = \frac{3\sqrt{3}}{14} + \frac{5\sqrt{3}}{14} + \frac{\sqrt{3}}{2} = \frac{8\sqrt{3}}{14} + \frac{7\sqrt{3}}{14} = \frac{15\sqrt{3}}{14} $$ 16. **Final answer:** $$ \sin A + \sin B + \sin C = \frac{15\sqrt{3}}{14} $$