1. **Problem Statement:** (a) Find the value of $x$ given the interior angles of a pentagon: $135^\circ$, $(2x)^\circ$, $(4x)^\circ$, $90^\circ$, and $85^\circ$. (b) Given a triangular field with stations A, B, and C: - Azimuth from A to B is $45^\circ$ and distance AB is 500 m. - Departure from B to C is 300 m and latitude from B to C is $-400$ m. Find: (i) Departure and latitude from B to A. (ii) Distance between B and C. (iii) Departure and latitude from A to C. --- 2. **Formulas and Rules:** - Sum of interior angles of a polygon with $n$ sides is $$(n-2) \times 180^\circ$$. - Departure and latitude relate to horizontal and vertical components of a vector. - Distance between two points given departure $d$ and latitude $l$ is $$\sqrt{d^2 + l^2}$$. - Azimuth angle $\theta$ relates to departure and latitude by: $$\text{departure} = \text{distance} \times \sin(\theta)$$ $$\text{latitude} = \text{distance} \times \cos(\theta)$$ --- 3. **Solution (a): Find $x$** Sum of interior angles of pentagon: $$\sum = (5-2) \times 180^\circ = 3 \times 180^\circ = 540^\circ$$ Sum given: $$135 + 2x + 4x + 90 + 85 = 540$$ Simplify: $$135 + 6x + 175 = 540$$ $$6x + 310 = 540$$ Subtract 310: $$6x = 540 - 310 = 230$$ Divide both sides by 6: $$x = \frac{230}{6}$$ Show cancellation: $$x = \frac{\cancel{230}}{\cancel{6}}$$ (no common factors, so simplified fraction is $\frac{230}{6}$) Simplify fraction: $$x = \frac{115}{3} \approx 38.33^\circ$$ --- 4. **Solution (b)(i): Departure and latitude from B to A** Given azimuth from A to B is $45^\circ$ and distance AB = 500 m. Departure from A to B: $$d_{AB} = 500 \times \sin 45^\circ = 500 \times \frac{\sqrt{2}}{2} = 353.55\,m$$ Latitude from A to B: $$l_{AB} = 500 \times \cos 45^\circ = 500 \times \frac{\sqrt{2}}{2} = 353.55\,m$$ Departure and latitude from B to A are negatives (reverse direction): $$d_{BA} = -d_{AB} = -353.55\,m$$ $$l_{BA} = -l_{AB} = -353.55\,m$$ --- 5. **Solution (b)(ii): Distance between B and C** Given departure $d_{BC} = 300$ m and latitude $l_{BC} = -400$ m. Distance: $$D_{BC} = \sqrt{300^2 + (-400)^2} = \sqrt{90000 + 160000} = \sqrt{250000} = 500\,m$$ --- 6. **Solution (b)(iii): Departure and latitude from A to C** Departure from A to C: $$d_{AC} = d_{AB} + d_{BC} = 353.55 + 300 = 653.55\,m$$ Latitude from A to C: $$l_{AC} = l_{AB} + l_{BC} = 353.55 + (-400) = -46.45\,m$$ --- **Final answers:** (a) $$x = \frac{115}{3} \approx 38.33^\circ$$ (b)(i) $$d_{BA} = -353.55\,m, \quad l_{BA} = -353.55\,m$$ (b)(ii) $$D_{BC} = 500\,m$$ (b)(iii) $$d_{AC} = 653.55\,m, \quad l_{AC} = -46.45\,m$$