1. **Problem 1: Distance the second ship must sail** A ship travels west 50 km, then sails 52.4 km in direction E25°N. We want the direct distance from the harbour to the ship's final position. 2. **Formula and approach:** We model the ship's path as two vectors: one 50 km west, the other 52.4 km at 25° north of east. We find the resultant vector's magnitude using the law of cosines or by breaking into components. 3. **Calculate components:** - West vector: $\vec{v_1} = (-50, 0)$ km (x west negative, y north positive) - Second vector components: $x_2 = 52.4 \cos 25^\circ$ $y_2 = 52.4 \sin 25^\circ$ Calculate: $\cos 25^\circ \approx 0.9063$, $\sin 25^\circ \approx 0.4226$ So, $x_2 = 52.4 \times 0.9063 = 47.48$ km $y_2 = 52.4 \times 0.4226 = 22.14$ km 4. **Resultant vector:** $x = -50 + 47.48 = -2.52$ km $y = 0 + 22.14 = 22.14$ km 5. **Distance from harbour:** $$d = \sqrt{(-2.52)^2 + (22.14)^2} = \sqrt{6.35 + 490.17} = \sqrt{496.52} \approx 22.3 \text{ km}$$ --- 6. **Problem 2a: Find angle $\theta$ in right triangle** Given sides opposite $\theta = 3.6$, adjacent $= 2.88$, hypotenuse $= 4.61$. 7. **Five methods to find $\theta$:** - Using sine: $\sin \theta = \frac{3.6}{4.61} \Rightarrow \theta = \sin^{-1}(0.7809) = 51^\circ$ - Using cosine: $\cos \theta = \frac{2.88}{4.61} \Rightarrow \theta = \cos^{-1}(0.6247) = 51^\circ$ - Using tangent: $\tan \theta = \frac{3.6}{2.88} \Rightarrow \theta = \tan^{-1}(1.25) = 51^\circ$ - Using Pythagorean theorem to verify hypotenuse: $\sqrt{3.6^2 + 2.88^2} = \sqrt{12.96 + 8.29} = \sqrt{21.25} = 4.61$ (confirms right triangle) - Using sine law (redundant here but possible): $\frac{3.6}{\sin \theta} = 4.61$ so $\theta = \sin^{-1}(3.6/4.61) = 51^\circ$ All methods give $\theta \approx 51^\circ$. --- 8. **Problem 2b: Choose method to solve for $\theta$** I would use tangent because it directly relates opposite and adjacent sides without needing the hypotenuse. --- 9. **Problem 3: Find side $d$ in two triangles with angles 64°, 82°, 34°** Sum of angles in triangle: $64 + 82 + 34 = 180^\circ$ confirms triangle. Use Law of Sines: $$\frac{d}{\sin 82^\circ} = \frac{2.12}{\sin 64^\circ}$$ Calculate: $\sin 82^\circ \approx 0.9903$, $\sin 64^\circ \approx 0.8988$ Solve for $d$: $$d = \frac{2.12 \times 0.9903}{0.8988} = 2.34 \text{ (rounded to nearest tenth)}$$ --- 10. **Problem 4: Find angle $\theta$ at roof peak** Given: - Width $AB = 15$ m - Wall height $AC = 3$ m - Roof sides $CD = DB = 12$ m Draw line $CB$ to split base. Triangle $CDB$ is isosceles with sides 12, 12, and base $CB$. Find $CB$ using Pythagoras in triangle $ACB$: $$CB = \sqrt{AB^2 + AC^2} = \sqrt{15^2 + 3^2} = \sqrt{225 + 9} = \sqrt{234} = 15.3$$ Use Law of Cosines in $\triangle CDB$: $$\cos \theta = \frac{12^2 + 12^2 - 15.3^2}{2 \times 12 \times 12} = \frac{144 + 144 - 234}{288} = \frac{54}{288} = 0.1875$$ $$\theta = \cos^{-1}(0.1875) = 79^\circ$$ --- 11. **Problem 5: Surveying problem** Measure one side (e.g., side $d$) and four angles (e.g., $\angle 1, \angle 2, \angle 3, \angle 4$). Use Law of Sines and Cosines to find other sides and angles: - Calculate side $a$ using Law of Sines: $$\frac{a}{\sin \angle 2} = \frac{d}{\sin \angle 1}$$ - Calculate side $b$ using Law of Cosines: $$b^2 = a^2 + c^2 - 2ac \cos \angle 3$$ - Calculate side $c$ similarly. - Calculate remaining angles by angle sum properties and Law of Sines. This approach uses minimal measurements and trigonometric relations to find all unknowns. --- **Final answers:** 1. Distance second ship must sail: $22.3$ km 2a. Angle $\theta \approx 51^\circ$ 3. Side $d \approx 2.3$ (nearest tenth) 4. Roof peak angle $\theta \approx 79^\circ$ 5. Surveying method described above.