1. **State the problem:** We need to find the value of $x$ for a triangle where the area is $\sqrt{5} \times 3$, the angle between the base and the side is $60^\circ$, the base length is $2x+1$, and the other side length (not the hypotenuse) is $x+2$. 2. **Formula for the area of a triangle using two sides and the included angle:** $$\text{Area} = \frac{1}{2} \times a \times b \times \sin(\theta)$$ where $a$ and $b$ are the lengths of two sides and $\theta$ is the angle between them. 3. **Apply the formula:** Given: - Area $= 3\sqrt{5}$ - $a = 2x + 1$ - $b = x + 2$ - $\theta = 60^\circ$ So, $$3\sqrt{5} = \frac{1}{2} \times (2x+1) \times (x+2) \times \sin(60^\circ)$$ 4. **Calculate $\sin(60^\circ)$:** $$\sin(60^\circ) = \frac{\sqrt{3}}{2}$$ 5. **Substitute $\sin(60^\circ)$ into the equation:** $$3\sqrt{5} = \frac{1}{2} \times (2x+1) \times (x+2) \times \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{4} (2x+1)(x+2)$$ 6. **Multiply both sides by 4 to clear the denominator:** $$4 \times 3\sqrt{5} = \sqrt{3} (2x+1)(x+2)$$ $$12\sqrt{5} = \sqrt{3} (2x+1)(x+2)$$ 7. **Divide both sides by $\sqrt{3}$:** $$\frac{12\sqrt{5}}{\sqrt{3}} = (2x+1)(x+2)$$ Simplify the left side: $$12 \times \sqrt{\frac{5}{3}} = (2x+1)(x+2)$$ 8. **Expand the right side:** $$(2x+1)(x+2) = 2x^2 + 4x + x + 2 = 2x^2 + 5x + 2$$ 9. **Set up the equation:** $$2x^2 + 5x + 2 = 12 \sqrt{\frac{5}{3}}$$ 10. **Rewrite as a quadratic equation:** $$2x^2 + 5x + 2 - 12 \sqrt{\frac{5}{3}} = 0$$ 11. **Use the quadratic formula:** $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $a=2$, $b=5$, and $c=2 - 12 \sqrt{\frac{5}{3}}$. Calculate discriminant: $$\Delta = 5^2 - 4 \times 2 \times \left(2 - 12 \sqrt{\frac{5}{3}}\right) = 25 - 8 + 96 \sqrt{\frac{5}{3}} = 17 + 96 \sqrt{\frac{5}{3}}$$ 12. **Final solutions:** $$x = \frac{-5 \pm \sqrt{17 + 96 \sqrt{\frac{5}{3}}}}{4}$$ This gives the possible values of $x$ that satisfy the problem.