1. **Problem statement:** Calculate angle CBD, the bearing of C from B, and the bearing of D from A given the quadrilateral ABCD with sides BC = 192 m, CD = 287.9 m, BD = 168 m, AD = 205.8 m, and bearing of D from B is 038°. 2. **(i) Calculate angle CBD:** We use the Law of Cosines in triangle BCD to find angle CBD. Formula: $$\cos(\angle CBD) = \frac{BD^2 + BC^2 - CD^2}{2 \times BD \times BC}$$ Substitute values: $$\cos(\angle CBD) = \frac{168^2 + 192^2 - 287.9^2}{2 \times 168 \times 192}$$ Calculate numerator: $$168^2 = 28224, \quad 192^2 = 36864, \quad 287.9^2 \approx 82845.61$$ So numerator: $$28224 + 36864 - 82845.61 = 65088 - 82845.61 = -17757.61$$ Calculate denominator: $$2 \times 168 \times 192 = 64512$$ Calculate cosine: $$\cos(\angle CBD) = \frac{-17757.61}{64512} \approx -0.2753$$ Find angle: $$\angle CBD = \cos^{-1}(-0.2753) \approx 106.0^\circ$$ This matches the required rounding. 3. **(ii) Find bearing of C from B:** Given bearing of D from B is 038°, and angle CBD = 106.0°, the bearing of C from B is: $$\text{Bearing of C from B} = 038^\circ + 106.0^\circ = 144.0^\circ$$ 4. **(iii) Calculate bearing of D from A:** Since A is due east of B, the bearing of B from A is 270° (west). We find angle ABD using Law of Cosines in triangle ABD: $$\cos(\angle ABD) = \frac{AB^2 + BD^2 - AD^2}{2 \times AB \times BD}$$ But AB is unknown. However, since A is due east of B, AB is horizontal distance. We can find AB using coordinates: Set B at origin (0,0), A at (x,0), so A is east of B. Coordinates of D from B using bearing 038° and distance 168 m: $$x_D = 168 \times \sin 38^\circ \approx 168 \times 0.6157 = 103.0$$ $$y_D = 168 \times \cos 38^\circ \approx 168 \times 0.7880 = 132.4$$ Coordinates of A are (AB,0), distance AD = 205.8 m: Distance formula: $$AD^2 = (x_D - AB)^2 + (y_D - 0)^2$$ Substitute: $$205.8^2 = (103.0 - AB)^2 + 132.4^2$$ Calculate squares: $$205.8^2 = 42349.64, \quad 132.4^2 = 17532.76$$ So: $$42349.64 = (103.0 - AB)^2 + 17532.76$$ Subtract: $$(103.0 - AB)^2 = 42349.64 - 17532.76 = 24816.88$$ Take square root: $$103.0 - AB = \pm 157.5$$ Two solutions: 1. $$103.0 - AB = 157.5 \Rightarrow AB = 103.0 - 157.5 = -54.5$$ (not possible since A is east of B) 2. $$103.0 - AB = -157.5 \Rightarrow AB = 103.0 + 157.5 = 260.5$$ So, $$AB = 260.5$$ m. Now, bearing of D from A is angle between line AD and north at A. Vector from A to D: $$x_D - x_A = 103.0 - 260.5 = -157.5$$ $$y_D - y_A = 132.4 - 0 = 132.4$$ Calculate angle from north: $$\theta = \tan^{-1} \left( \frac{|x|}{y} \right) = \tan^{-1} \left( \frac{157.5}{132.4} \right) \approx \tan^{-1}(1.19) = 50.2^\circ$$ Since x is negative and y positive, point D is northwest of A, so bearing is: $$360^\circ - 50.2^\circ = 309.8^\circ$$ **Final answers:** (i) Angle CBD = 106.0° (ii) Bearing of C from B = 144.0° (iii) Bearing of D from A = 309.8°