1. **State the problem:** We need to find the sum of the vectors $\langle 1, -2 \rangle$ and $\langle 1, 8 \rangle$, then find the magnitude and direction of the resultant vector. Round angles to the nearest degree and other values to the nearest tenth. 2. **Add the vectors:** $$\langle 1, -2 \rangle + \langle 1, 8 \rangle = \langle 1+1, -2+8 \rangle = \langle 2, 6 \rangle$$ 3. **Find the magnitude of the resultant vector:** The magnitude formula is: $$\text{magnitude} = \sqrt{x^2 + y^2}$$ where $x=2$ and $y=6$. Calculate: $$\sqrt{2^2 + 6^2} = \sqrt{4 + 36} = \sqrt{40}$$ Simplify: $$\sqrt{40} = \sqrt{4 \times 10} = 2\sqrt{10} \approx 6.3$$ 4. **Find the direction (angle) of the resultant vector:** The direction $\theta$ is given by: $$\theta = \tan^{-1}\left(\frac{y}{x}\right) = \tan^{-1}\left(\frac{6}{2}\right) = \tan^{-1}(3)$$ Calculate: $$\theta \approx 71.6^\circ$$ Rounded to the nearest degree: $$72^\circ$$ 5. **Summary:** - Resultant vector: $\langle 2, 6 \rangle$ - Magnitude: $6.3$ - Direction: $72^\circ$ **Note:** None of the provided options match this result exactly, so the correct resultant vector and its magnitude and direction are as calculated above.