1. **State 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. **Use the distance formula:** The distance $OP$ is given by $$OP = \sqrt{k^2 + 5^2} = 13.$$ 3. **Set up the equation:** $$\sqrt{k^2 + 25} = 13.$$ 4. **Square both sides to eliminate the square root:** $$k^2 + 25 = 13^2 = 169.$$ 5. **Solve for $k^2$:** $$k^2 = 169 - 25 = 144.$$ 6. **Find $k$:** Since $P$ is in the second quadrant, $k$ (the $x$-coordinate) is negative: $$k = -\sqrt{144} = -12.$$ --- 7. **Calculate $26 \cos \theta$ without a calculator:** Recall that $\cos \theta = \frac{\text{adjacent side}}{\text{hypotenuse}} = \frac{k}{13}.$ 8. **Substitute $k = -12$:** $$\cos \theta = \frac{-12}{13}.$$ 9. **Calculate $26 \cos \theta$:** $$26 \cos \theta = 26 \times \frac{-12}{13} = 2 \times (-12) = -24.$$ **Final answers:** - $k = -12$ - $26 \cos \theta = -24$