1. **Problem statement:** We have quadrilateral ABCD with given sides and angles: CD = 10 m, DB = 12 m, angle DBA = 90°, angle CDB = 56°, and angle ADB = 34°. We need to find (a) the length of AB and (b) the area of ABCD. 2. **Step (a) Calculate length AB:** - Triangle DBA is right-angled at B. - Use the Law of Sines in triangle DBA or trigonometry since angle DBA = 90°. - In triangle DBA, angle ADB = 34°, angle DBA = 90°, so angle BAD = 56° (since angles sum to 180°). - DB = 12 m is the hypotenuse of right triangle DBA. - Using sine for side AB opposite angle ADB: $$AB = DB \times \sin(34^\circ)$$ - Calculate: $$AB = 12 \times \sin(34^\circ) \approx 12 \times 0.5592 = 6.7104$$ - So, $$AB \approx 6.7\,m$$ 3. **Step (b) Calculate area of quadrilateral ABCD:** - Quadrilateral ABCD can be split into two triangles: CDB and DBA. - **Area of triangle CDB:** - Use formula $$\text{Area} = \frac{1}{2} \times CD \times DB \times \sin(\angle CDB)$$ - Given CD = 10 m, DB = 12 m, angle CDB = 56° - Calculate: $$\text{Area}_{CDB} = \frac{1}{2} \times 10 \times 12 \times \sin(56^\circ)$$ $$= 60 \times 0.8290 = 49.74$$ - **Area of triangle DBA:** - Right triangle with legs AB and DB perpendicular at B. - Area formula for right triangle: $$\text{Area}_{DBA} = \frac{1}{2} \times AB \times DB \times \cos(56^\circ)$$ - But since angle DBA = 90°, and AB is opposite 34°, better to use: $$\text{Area}_{DBA} = \frac{1}{2} \times AB \times BD \times \cos(56^\circ)$$ - Alternatively, use: $$\text{Area}_{DBA} = \frac{1}{2} \times AB \times BD \times \sin(90^\circ) = \frac{1}{2} \times AB \times BD$$ - Since angle DBA = 90°, area is: $$\text{Area}_{DBA} = \frac{1}{2} \times AB \times BD = \frac{1}{2} \times 6.7 \times 12 = 40.2$$ - **Total area of ABCD:** $$\text{Area}_{ABCD} = \text{Area}_{CDB} + \text{Area}_{DBA} = 49.74 + 40.2 = 89.94$$ - Rounded: $$\text{Area}_{ABCD} \approx 90.0\,m^2$$ **Final answers:** - (a) $$AB \approx 6.7\,m$$ - (b) $$\text{Area} \approx 90.0\,m^2$$