1. **Problem statement:** Find $\sec \theta$ for the angle $\alpha$ in the right triangle with sides $RP=18$, $PQ=27$, and hypotenuse $RQ$. 2. **Recall the definition:** $\sec \theta = \frac{1}{\cos \theta}$ and $\cos \theta = \frac{\text{adjacent side}}{\text{hypotenuse}}$. 3. **Identify sides relative to angle $\alpha$:** - Adjacent side to $\alpha$ is $RP = 18$. - Opposite side to $\alpha$ is $PQ = 27$. - Hypotenuse $RQ$ is unknown. 4. **Calculate hypotenuse using Pythagoras theorem:** $$RQ = \sqrt{RP^2 + PQ^2} = \sqrt{18^2 + 27^2} = \sqrt{324 + 729} = \sqrt{1053}$$ 5. **Simplify $\sqrt{1053}$:** $$1053 = 9 \times 117 = 9 \times 9 \times 13 = 81 \times 13$$ $$\Rightarrow \sqrt{1053} = \sqrt{81 \times 13} = 9 \sqrt{13}$$ 6. **Calculate $\cos \alpha$:** $$\cos \alpha = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{18}{9 \sqrt{13}}$$ 7. **Simplify the fraction:** $$\frac{18}{9 \sqrt{13}} = \frac{\cancel{18}^{2}}{\cancel{9}^{1} \sqrt{13}} = \frac{2}{\sqrt{13}}$$ 8. **Calculate $\sec \alpha$:** $$\sec \alpha = \frac{1}{\cos \alpha} = \frac{1}{\frac{2}{\sqrt{13}}} = \frac{\sqrt{13}}{2}$$ **Final answer:** $$\boxed{\sec \alpha = \frac{\sqrt{13}}{2}}$$