1. **Problem c)** Solve for $\theta$ given $\tan \theta = \frac{1}{\sqrt{3}}$ for $-360 \leq \theta \leq 360$. - Recall that $\tan \theta = \frac{1}{\sqrt{3}}$ corresponds to angles where tangent is positive and equals $\frac{1}{\sqrt{3}}$. - The reference angle is $\theta_r = 30^\circ$ because $\tan 30^\circ = \frac{1}{\sqrt{3}}$. - Tangent is positive in Quadrants I and III. - Within $-360^\circ$ to $360^\circ$, the solutions are: - Quadrant I: $30^\circ$, $30^\circ - 360^\circ = -330^\circ$ - Quadrant III: $180^\circ + 30^\circ = 210^\circ$, $210^\circ - 360^\circ = -150^\circ$ **Final solutions:** $\theta = -330^\circ, -150^\circ, 30^\circ, 210^\circ$. 2. **Problem 5(i)** Find distance $AC$ given yacht travels from $A$ to $B$ (5 km, bearing $067^\circ$), then from $B$ to $C$ (8 km, bearing $146^\circ$). - Convert bearings to standard angles from the positive x-axis: - Bearing $067^\circ$ means $67^\circ$ from north clockwise, so angle from east is $90^\circ - 67^\circ = 23^\circ$. - Bearing $146^\circ$ means $146^\circ$ from north clockwise, angle from east is $90^\circ - 146^\circ = -56^\circ$. - Calculate coordinates: - Point $B$: $x_B = 5 \cos 23^\circ$, $y_B = 5 \sin 23^\circ$ - Point $C$: $x_C = x_B + 8 \cos (-56^\circ)$, $y_C = y_B + 8 \sin (-56^\circ)$ - Compute: - $x_B \approx 5 \times 0.9205 = 4.6025$ - $y_B \approx 5 \times 0.3907 = 1.9535$ - $x_C \approx 4.6025 + 8 \times 0.5592 = 4.6025 + 4.4736 = 9.0761$ - $y_C \approx 1.9535 + 8 \times (-0.8290) = 1.9535 - 6.632 = -4.6785$ - Distance $AC = \sqrt{(x_C - 0)^2 + (y_C - 0)^2} = \sqrt{9.0761^2 + (-4.6785)^2} \approx \sqrt{82.38 + 21.89} = \sqrt{104.27} \approx 10.21$ km. 3. **Problem 5(ii)** Find bearing of $C$ from $A$. - Bearing is angle clockwise from north to line $AC$. - Calculate angle $\alpha$ from east axis: $\alpha = \arctan \left( \frac{y_C}{x_C} \right) = \arctan \left( \frac{-4.6785}{9.0761} \right) \approx -27.1^\circ$. - Since $x_C > 0$ and $y_C < 0$, point $C$ is in Quadrant IV. - Bearing from north is $90^\circ - \alpha = 90^\circ - (-27.1^\circ) = 117.1^\circ$. **Final bearing:** approximately $117^\circ$. 4. **Problem 3** Solve $2 \sin^2 \theta - \cos \theta - 1 = 0$ for $0 \leq \theta \leq 360$. - Use identity $\sin^2 \theta = 1 - \cos^2 \theta$: $$2(1 - \cos^2 \theta) - \cos \theta - 1 = 0$$ $$2 - 2 \cos^2 \theta - \cos \theta - 1 = 0$$ $$-2 \cos^2 \theta - \cos \theta + 1 = 0$$ Multiply both sides by $-1$: $$2 \cos^2 \theta + \cos \theta - 1 = 0$$ - Let $x = \cos \theta$, then: $$2x^2 + x - 1 = 0$$ - Solve quadratic: $$x = \frac{-1 \pm \sqrt{1^2 - 4 \times 2 \times (-1)}}{2 \times 2} = \frac{-1 \pm \sqrt{1 + 8}}{4} = \frac{-1 \pm 3}{4}$$ - Solutions: - $x = \frac{-1 + 3}{4} = \frac{2}{4} = 0.5$ - $x = \frac{-1 - 3}{4} = \frac{-4}{4} = -1$ - Find $\theta$: - For $\cos \theta = 0.5$, $\theta = 60^\circ$ or $300^\circ$. - For $\cos \theta = -1$, $\theta = 180^\circ$. **Final solutions:** $\theta = 60^\circ, 180^\circ, 300^\circ$.