Question:

Step 1: Understand the given problem and identify the key information provided. The problem gives us $\cos(\theta + 70°) = \dfrac{-1}{3}$, where $\theta$ is an angle between $0°$ and $110°$. This implies that $\theta + 70°$ falls in the second quadrant, as the sum of $\theta$ and $70°$ ranges between $70°$ and $180°$. Since $\theta + 70°$ is in the second quadrant, $\sin(\theta + 70°)$ will be positive. Step 2: Calculate $\sin(\theta + 70°)$ using the given $\cos(\theta + 70°)$. We use the trigonometric identity $\sin^2(\alpha) + \cos^2(\alpha) = 1$ to find $\sin(\theta + 70°)$. Given $\cos(\theta + 70°) = \dfrac{-1}{3}$, we have $\sin(\theta + 70°) = \sqrt{1 - \cos^2(\theta + 70°)} = \sqrt{1 - \left(\dfrac{-1}{3}\right)^2} = \sqrt{1 - \dfrac{1}{9}} = \sqrt{\dfrac{8}{9}} = \dfrac{2\sqrt{2}}{3}$. Step 3: Calculate $\tan(\theta + 70°)$ using the values of $\sin(\theta + 70°)$ and $\cos(\theta + 70°)$. The tangent of an angle is given by $\tan(\alpha) = \dfrac{\sin(\alpha)}{\cos(\alpha)}$. Substituting the known values, we get $\tan(\theta + 70°) = \dfrac{\sin(\theta + 70°)}{\cos(\theta + 70°)} = \dfrac{\dfrac{2\sqrt{2}}{3}}{-\dfrac{1}{3}} = -2\sqrt{2}$. Step 4: Calculate $\cos(160° + \theta)$ using the angle addition formula. We recognize that $\cos(160° + \theta) = \cos((\theta + 70°) + 90°)$. Using the angle addition formula for cosine, which states $\cos(\alpha + \beta) = \cos(\alpha)\cos(\beta) - \sin(\alpha)\sin(\beta)$, and knowing that $\cos(90°) = 0$ and $\sin(90°) = 1$, we simplify to $\cos(160° + \theta) = -\sin(\theta + 70°) = -\dfrac{2\sqrt{2}}{3}$. Step 5: Calculate $\sin(20° - \theta)$ using the angle subtraction formula. Recognizing that $\sin(20° - \theta) = -\sin(\theta - 20°)$ and further simplifying using $\sin(\theta - 20°) = \sin((\theta + 70°) - 90°)$, we apply the angle subtraction formula for sine, which yields $-\sin(\theta + 70° - 90°)$. Since $\sin(\alpha - 90°) = -\cos(\alpha)$, we have $-\sin(\theta + 70° - 90°) = -(-\cos(\theta + 70°)) = \cos(\theta + 70°) = -\dfrac{1}{3}$. Step 6: Calculate $\tan(25° + \theta)$ using the tangent subtraction formula. Noting that $25° + \theta = (\theta + 70°) - 45°$, we apply the tangent subtraction formula: $\tan(\alpha - \beta) = \dfrac{\tan(\alpha) - \tan(\beta)}{1 + \tan(\alpha)\tan(\beta)}$. Substituting the known values, $\tan(25° + \theta) = \dfrac{\tan(\theta + 70°) - \tan(45°)}{1 + \tan(\theta + 70°)\tan(45°)} = \dfrac{-2\sqrt{2} - 1}{1 + (-2\sqrt{2}) \cdot 1} = \dfrac{-(2\sqrt{2} + 1)}{1 - 2\sqrt{2}}$. Step 7: Rationalize the denominator of $\tan(25° + \theta)$. To rationalize the denominator, we multiply both the numerator and the denominator by the conjugate of the denominator, $1 + 2\sqrt{2}$. This gives $\dfrac{-(2\sqrt{2} + 1)(1 + 2\sqrt{2})}{(1 - 2\sqrt{2})(1 + 2\sqrt{2})} = \dfrac{-(2\sqrt{2} + 1)^2}{1 - 8} = \dfrac{-(8 + 4\sqrt{2} + 1)}{-7} = \dfrac{9 + 4\sqrt{2}}{7}$. Step 8: Conclude the calculations and match the results with the given options. From the calculations, we have $P \to 3$, $Q \to 4$, $R \to 5$, and $S \to 2$. Therefore, the correct answer is option 2.