1) Solve the trigonometric equations within the given intervals. **a) Solve $\sin \theta = -\frac{1}{2}$ for $-360 \leq \theta \leq 360$** 1. The sine function equals $-\frac{1}{2}$ at reference angle $30^\circ$ (or $\frac{\pi}{6}$ radians). 2. Sine is negative in the third and fourth quadrants. 3. Solutions in $[0,360]$ are $\theta = 210^\circ$ and $330^\circ$. 4. Considering the interval $-360 \leq \theta \leq 360$, include negative coterminal angles: - $210^\circ - 360^\circ = -150^\circ$ - $330^\circ - 360^\circ = -30^\circ$ 5. Final solutions: $$\theta = -150^\circ, -30^\circ, 210^\circ, 330^\circ$$ **b) Solve $\cos \theta = -1$ for $-260 \leq \theta \leq 360$** 1. Cosine equals $-1$ at $180^\circ$ plus multiples of $360^\circ$. 2. Within $-260 \leq \theta \leq 360$, check: - $180^\circ$ is in the interval. - $180^\circ - 360^\circ = -180^\circ$ is also in the interval. 3. Final solutions: $$\theta = -180^\circ, 180^\circ$$ **c) Solve $\tan \theta = \frac{1}{\sqrt{3}}$ for $-360 \leq \theta \leq 360$** 1. $\tan \theta = \frac{1}{\sqrt{3}}$ at $\theta = 30^\circ$ plus $180^\circ k$, where $k$ is any integer. 2. Find all solutions in the interval: - $30^\circ$ - $30^\circ + 180^\circ = 210^\circ$ - $30^\circ - 180^\circ = -150^\circ$ - $30^\circ - 360^\circ = -330^\circ$ 3. Final solutions: $$\theta = -330^\circ, -150^\circ, 30^\circ, 210^\circ$$ --- 2) Yacht bearing problem: Given: - Vector $\overrightarrow{AB}$: length 5 km, bearing $67^\circ$ - Vector $\overrightarrow{BC}$: length 8 km, bearing $146^\circ$ (i) Find distance $AC$. 1. Convert bearings to standard angles from the positive x-axis (East): - Bearing $67^\circ$ means $67^\circ$ clockwise from North. - In standard position, angle from East is $90^\circ - 67^\circ = 23^\circ$. - Bearing $146^\circ$ means $146^\circ$ clockwise from North. - Standard angle from East is $90^\circ - 146^\circ = -56^\circ$ (or $304^\circ$). 2. Find coordinates of points: - $A$ at origin $(0,0)$. - $B$ at: $$x_B = 5 \cos 23^\circ, \quad y_B = 5 \sin 23^\circ$$ - $C$ at: $$x_C = x_B + 8 \cos (-56^\circ), \quad y_C = y_B + 8 \sin (-56^\circ)$$ 3. Calculate: $$x_B = 5 \times 0.9205 = 4.6025$$ $$y_B = 5 \times 0.3907 = 1.9535$$ $$x_C = 4.6025 + 8 \times 0.5592 = 4.6025 + 4.4736 = 9.0761$$ $$y_C = 1.9535 + 8 \times (-0.8290) = 1.9535 - 6.632 = -4.6785$$ 4. Distance $AC$: $$AC = \sqrt{(9.0761 - 0)^2 + (-4.6785 - 0)^2} = \sqrt{82.38 + 21.89} = \sqrt{104.27} \approx 10.21 \text{ km}$$ (ii) Find bearing of $C$ from $A$. 1. Calculate angle from East: $$\theta = \arctan \left( \frac{y_C}{x_C} \right) = \arctan \left( \frac{-4.6785}{9.0761} \right) = -27.1^\circ$$ 2. Convert to bearing from North: $$\text{Bearing} = 90^\circ - (-27.1^\circ) = 117.1^\circ$$ 3. Since $x_C > 0$ and $y_C < 0$, point $C$ is in the fourth quadrant, so bearing is correct. **Final answers:** - $AC \approx 10.21$ km - Bearing of $C$ from $A$ is approximately $117^\circ$