1. **Problem statement:** We have four statues A, B, C, and D with given angles and lengths. We need to find angles ACD, BCD, ACB, and length AB. 2. **Given data:** - $\angle BAC = \angle ADC = 90^\circ$ - $AC = 10$ cm - $AD = 9.4$ cm - Bearing of D from C is $242^\circ$ - C is due east of B - $\angle ACD$ is to be found and shown as $70^\circ$ (nearest degree) 3. **Step (i): Find $\angle ACD$ in triangle ACD** - Triangle ACD has right angle at D ($\angle ADC = 90^\circ$) - Use Pythagoras or trigonometry to find $\angle ACD$ - Using cosine rule or trigonometric tables: $$\cos(\angle ACD) = \frac{AD}{AC} = \frac{9.4}{10} = 0.94$$ - From trigonometric tables or calculator, $\cos^{-1}(0.94) \approx 20^\circ$ - Since $\angle ADC = 90^\circ$, $\angle ACD = 70^\circ$ (because angles in triangle sum to 180°) 4. **Step (ii): Find $\angle BCD$ and $\angle ACB$** - Bearing of D from C is $242^\circ$. - Since C is due east of B, line BC is horizontal pointing west to east. - Bearing $242^\circ$ means angle from north clockwise to line CD is $242^\circ$. - Angle between BC (east) and CD is $242^\circ - 180^\circ = 62^\circ$ (since east is 90°, south is 180°, west is 270°) - So, $\angle BCD = 62^\circ$ - Triangle ABC has right angle at C ($\angle BAC = 90^\circ$) - Sum of angles in triangle ABC: $90^\circ + \angle ACB + \angle BCD = 180^\circ$ - So, $\angle ACB = 180^\circ - 90^\circ - 62^\circ = 28^\circ$ 5. **Step (iii): Find length AB in triangle ABC** - Triangle ABC right angled at C - Use sine rule or trigonometry: $$\sin(\angle ACB) = \frac{BC}{AB}$$ - We know $AC = 10$ cm, $\angle ACB = 28^\circ$, and $\angle BCD = 62^\circ$ - Since C is east of B, $BC = AC \times \tan(\angle ACB) = 10 \times \tan(28^\circ) \approx 10 \times 0.5317 = 5.317$ cm - Use Pythagoras theorem: $$AB = \sqrt{AC^2 + BC^2} = \sqrt{10^2 + 5.317^2} = \sqrt{100 + 28.28} = \sqrt{128.28} \approx 11.33 \text{ cm}$$ **Final answers:** - $\angle ACD = 70^\circ$ - $\angle BCD = 62^\circ$ - $\angle ACB = 28^\circ$ - $AB \approx 11.33$ cm