1. **Stating the problem:** We have a point $P(k, 5)$ in the second quadrant such that the distance from the origin $O$ to $P$ is 13 units. We need to find the value of $k$. 2. **Using the distance formula:** The distance $OP$ is given by $$OP = \sqrt{k^2 + 5^2} = 13$$ 3. **Solving for $k$:** Square both sides: $$k^2 + 25 = 169$$ $$k^2 = 169 - 25 = 144$$ 4. **Finding $k$:** $$k = \pm \sqrt{144} = \pm 12$$ 5. **Considering the quadrant:** Since $P$ is in the second quadrant, $k$ (the $x$-coordinate) must be negative. Therefore, $$k = -12$$ --- 6. **Finding $\cos \theta$:** Angle $\theta$ is formed at $O$ between $OP$ and $OQ$, where $Q$ lies on the positive $x$-axis. 7. **Using the definition of cosine:** $$\cos \theta = \frac{\text{adjacent side}}{\text{hypotenuse}} = \frac{k}{OP} = \frac{-12}{13}$$ 8. **Calculating $26 \cos \theta$:** $$26 \cos \theta = 26 \times \frac{-12}{13} = \cancel{26}^2 \times \frac{-12}{\cancel{13}^1} = 2 \times (-12) = -24$$ **Final answers:** - $k = -12$ - $26 \cos \theta = -24$