1. Problem 5: A ship sails 20 km due north and then 35 km due east. Find how far it is from its starting point. Step 1: Identify the path of the ship forms a right triangle with legs 20 km (north) and 35 km (east). Step 2: Use the Pythagorean theorem to find the distance from the starting point: $$d = \sqrt{20^2 + 35^2}$$. Step 3: Calculate the value: $$d = \sqrt{400 + 1225} = \sqrt{1625} \approx 40.31 \text{ km}$$. Answer: The ship is approximately 40.31 km from its starting point. 2. Problem 6: Aircraft flies 400 km at bearing 025°, then 700 km at bearing 080° to arrive at B. Find: a) How far north of O is B? Step 1: Resolve each leg into north (y) and east (x) components. Step 2: For 400 km at 025°, $$x_1 = 400 \sin 25^\circ$$ $$y_1 = 400 \cos 25^\circ$$ Step 3: For 700 km at 080°, $$x_2 = 700 \sin 80^\circ$$ $$y_2 = 700 \cos 80^\circ$$ Step 4: Total north distance: $$y = y_1 + y_2 = 400 \cos 25^\circ + 700 \cos 80^\circ$$ Calculate: $$400 \cos 25^\circ \approx 400 \times 0.9063 = 362.52$$ $$700 \cos 80^\circ \approx 700 \times 0.1736 = 121.52$$ $$y = 362.52 + 121.52 = 484.04 \text{ km}$$ Answer a): 484.04 km north of O. b) How far east of O is B? Step 5: Total east distance: $$x = x_1 + x_2 = 400 \sin 25^\circ + 700 \sin 80^\circ$$ Calculate: $$400 \sin 25^\circ \approx 400 \times 0.4226 = 169.04$$ $$700 \sin 80^\circ \approx 700 \times 0.9848 = 689.36$$ $$x = 169.04 + 689.36 = 858.40 \text{ km}$$ Answer b): 858.40 km east of O. c) Find the distance and bearing of B from O. Step 6: Distance from O to B: $$d = \sqrt{x^2 + y^2} = \sqrt{858.40^2 + 484.04^2}$$ Calculate: $$d \approx \sqrt{736872 + 234298} = \sqrt{971170} = 985.49 \text{ km}$$ Step 7: Bearing = angle measured clockwise from north: $$\theta = \tan^{-1} \left( \frac{x}{y} \right) = \tan^{-1} \left( \frac{858.40}{484.04} \right)$$ Calculate: $$\theta \approx \tan^{-1} (1.774) = 60.4^\circ$$ Bearing = 60.4° Answer c): Distance is 985.49 km on a bearing of 060.4° from O. 3. Problem 7: Aircraft flies 500 km bearing 100°, then 600 km bearing 160°. Find distance and bearing from start. Step 1: Resolve vectors. For 500 km, bearing 100°: $$x_1 = 500 \sin 100^\circ$$ $$y_1 = 500 \cos 100^\circ$$ Compute: $$x_1 = 500 \times 0.9848 = 492.40$$ $$y_1 = 500 \times -0.1736 = -86.80$$ For 600 km, bearing 160°: $$x_2 = 600 \sin 160^\circ$$ $$y_2 = 600 \cos 160^\circ$$ Compute: $$x_2 = 600 \times 0.3420 = 205.20$$ $$y_2 = 600 \times -0.9397 = -563.82$$ Step 2: Total displacement components: $$x = 492.40 + 205.20 = 697.60$$ $$y = -86.80 + (-563.82) = -650.62$$ Step 3: Distance: $$d = \sqrt{697.60^2 + (-650.62)^2} = \sqrt{486648 + 423306} = \sqrt{909954} \approx 953.39 \text{ km}$$ Step 4: Bearing, angle with north: $$\theta = \tan^{-1} \left( \frac{|x|}{|y|} \right) = \tan^{-1} \left( \frac{697.60}{650.62} \right) = 46.7^\circ$$ Since y is negative (south) and x positive (east), bearing is $$180^\circ - 46.7^\circ = 133.3^\circ$$ Answer: Distance approx 953.39 km on bearing 133.3°. 4. Problem 8: A ship sails 95 km bearing 140°, then 102 km bearing 260°, then returns directly to start. Find distance and bearing of return journey. Step 1: Calculate components for first leg: $$x_1 = 95 \sin 140^\circ = 95 \times 0.6428 = 61.07$$ $$y_1 = 95 \cos 140^\circ = 95 \times -0.7660 = -72.77$$ Step 2: Calculate components for second leg: $$x_2 = 102 \sin 260^\circ = 102 \times -0.9848 = -100.46$$ $$y_2 = 102 \cos 260^\circ = 102 \times -0.1736 = -17.71$$ Step 3: Sum position after two legs: $$x = 61.07 + (-100.46) = -39.39$$ $$y = -72.77 + (-17.71) = -90.48$$ Step 4: Return journey vector is opposite of (x,y), so components: $$x_r = 39.39$$ $$y_r = 90.48$$ Step 5: Distance of return journey: $$d = \sqrt{39.39^2 + 90.48^2} = \sqrt{1551.6 + 8186.6} = \sqrt{9738.2} \approx 98.68 \text{ km}$$ Step 6: Bearing of return journey: $$\theta = \tan^{-1} \left( \frac{39.39}{90.48} \right) = 23.4^\circ$$ Since both components positive, bearing is 23.4°. Answer: Return journey distance is approx 98.68 km on a bearing of 023.4°.