1. **State the problem:** We have two right-angled triangles side by side with angles $p$ and $r$ such that $\sin p = \sin r$. We need to show that this implies the quadratic equation $$12x^2 - 17x - 5 = 0$$ and then calculate the size of angle $p$. 2. **Recall the sine formula for right triangles:** For a right triangle, $$\sin(\text{angle}) = \frac{\text{opposite side}}{\text{hypotenuse}}$$. 3. **Express $\sin p$ and $\sin r$ using given sides:** $$\sin p = \frac{6x + 5}{12x + 31}$$ $$\sin r = \frac{x}{4x - 1}$$ 4. **Set $\sin p = \sin r$ and form an equation:** $$\frac{6x + 5}{12x + 31} = \frac{x}{4x - 1}$$ 5. **Cross multiply:** $$(6x + 5)(4x - 1) = x(12x + 31)$$ 6. **Expand both sides:** $$24x^2 - 6x + 20x - 5 = 12x^2 + 31x$$ $$24x^2 + 14x - 5 = 12x^2 + 31x$$ 7. **Bring all terms to one side:** $$24x^2 + 14x - 5 - 12x^2 - 31x = 0$$ $$12x^2 - 17x - 5 = 0$$ This proves part (a). 8. **Solve the quadratic equation:** Use the quadratic formula: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $a=12$, $b=-17$, $c=-5$. Calculate discriminant: $$\Delta = (-17)^2 - 4 \times 12 \times (-5) = 289 + 240 = 529$$ Square root: $$\sqrt{529} = 23$$ Calculate roots: $$x = \frac{17 \pm 23}{24}$$ Two solutions: $$x_1 = \frac{17 + 23}{24} = \frac{40}{24} = \frac{5}{3}$$ $$x_2 = \frac{17 - 23}{24} = \frac{-6}{24} = -\frac{1}{4}$$ Since side lengths must be positive, discard $x = -\frac{1}{4}$. 9. **Calculate angle $p$ using $x=\frac{5}{3}$:** $$\sin p = \frac{6x + 5}{12x + 31} = \frac{6 \times \frac{5}{3} + 5}{12 \times \frac{5}{3} + 31} = \frac{10 + 5}{20 + 31} = \frac{15}{51} = \frac{5}{17}$$ 10. **Find angle $p$:** $$p = \sin^{-1}\left(\frac{5}{17}\right) \approx 17.1^\circ$$ **Final answer:** $$12x^2 - 17x - 5 = 0$$ $$p \approx 17.1^\circ$$