1. **State the problem:** We have triangle ABC with point D on AC such that angle BDC = 90°. Given: $BD=10$ cm, $AB=15$ cm, $DC=12.5$ cm. We need to find: (a) Length $AD$ correct to 3 significant figures. (b) Size of angle $BCD$ correct to 1 decimal place. 2. **Use the Pythagorean theorem in triangle BDC:** Since $\angle BDC = 90^\circ$, triangle BDC is right-angled at D. 3. **Calculate length BC:** $$BC = \sqrt{BD^2 + DC^2} = \sqrt{10^2 + 12.5^2} = \sqrt{100 + 156.25} = \sqrt{256.25} = 16.0\text{ cm}$$ 4. **Use the Law of Cosines in triangle ABC to find AD:** We know $AB=15$ cm, $BC=16.0$ cm, and $DC=12.5$ cm. Since $AC = AD + DC$, let $AD = x$. Using the Pythagorean theorem in triangle ABD (right angled at D), $$AB^2 = AD^2 + BD^2$$ $$15^2 = x^2 + 10^2$$ $$225 = x^2 + 100$$ $$x^2 = 125$$ $$x = \sqrt{125} = 11.1803\text{ cm}$$ Rounded to 3 significant figures: $$AD = 11.2\text{ cm}$$ 5. **Calculate angle BCD:** In triangle BCD, use tangent: $$\tan(\theta) = \frac{BD}{DC} = \frac{10}{12.5} = 0.8$$ Calculate $\theta$: $$\theta = \tan^{-1}(0.8) = 38.66^\circ$$ Rounded to 1 decimal place: $$\theta = 38.7^\circ$$ **Final answers:** (a) $AD = 11.2$ cm (b) $\angle BCD = 38.7^\circ$