1. **Problem statement:** Given a right triangle with points A, B, C, and D such that triangle ABC is right-angled at A, with sides AB = 3, BC = 5, BD = \sqrt{11}, and DC = 6. 2. **Show that AC = 4:** Since ABC is right-angled at A, by the Pythagorean theorem: $$AB^2 + AC^2 = BC^2$$ Substitute the known values: $$3^2 + AC^2 = 5^2$$ $$9 + AC^2 = 25$$ $$AC^2 = 25 - 9 = 16$$ $$AC = \sqrt{16} = 4$$ 3. **Show that triangle BCD is right-angled at B:** Check if the Pythagorean theorem holds for triangle BCD with sides BD = \sqrt{11}, DC = 6, and BC = 5. Calculate: $$BD^2 + BC^2 = (\sqrt{11})^2 + 5^2 = 11 + 25 = 36$$ Compare with: $$DC^2 = 6^2 = 36$$ Since $$BD^2 + BC^2 = DC^2$$, triangle BCD is right-angled at B. 4. **Calculate the trigonometric ratios for angle AĈB:** Angle AĈB is angle at point C between points A and B. From triangle ABC: - Opposite side to angle AĈB is AB = 3 - Adjacent side to angle AĈB is AC = 4 - Hypotenuse is BC = 5 Calculate: $$\tan A\hat{C}B = \frac{\text{opposite}}{\text{adjacent}} = \frac{AB}{AC} = \frac{3}{4}$$ $$\sin A\hat{C}B = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{AB}{BC} = \frac{3}{5}$$ **Final answers:** - $AC = 4$ - Triangle BCD is right-angled at B - $\tan A\hat{C}B = \frac{3}{4}$ - $\sin A\hat{C}B = \frac{3}{5}$