1. **Problem statement:** Find the hypotenuse $RQ$ of right triangle $QRP$ with right angle at $P$, sides $RP=18$ and $QP=27$. 2. **Formula:** Use the Pythagorean theorem: $$RQ^2 = RP^2 + QP^2$$ 3. **Calculate:** $$RQ^2 = 18^2 + 27^2 = 324 + 729 = 1053$$ 4. **Simplify:** $$RQ = \sqrt{1053} = \sqrt{9 \times 117} = 3\sqrt{117} = 3\sqrt{9 \times 13} = 3 \times 3 \sqrt{13} = 9\sqrt{13}$$ **Final answer:** The hypotenuse $RQ$ is $9\sqrt{13}$. --- **Additional trigonometric values for angle $\alpha$ at vertex $R$:** - $\sin \alpha = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{27}{9\sqrt{13}} = \frac{3}{\sqrt{13}} = \frac{3\sqrt{13}}{13}$ - $\cos \alpha = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{18}{9\sqrt{13}} = \frac{2}{\sqrt{13}} = \frac{2\sqrt{13}}{13}$ - $\tan \alpha = \frac{\text{opposite}}{\text{adjacent}} = \frac{27}{18} = \frac{3}{2}$ - $\csc \alpha = \frac{1}{\sin \alpha} = \frac{9\sqrt{13}}{27} = \frac{\sqrt{13}}{3}$ - $\sec \alpha = \frac{1}{\cos \alpha} = \frac{9\sqrt{13}}{18} = \frac{\sqrt{13}}{2}$ - $\cot \alpha = \frac{1}{\tan \alpha} = \frac{18}{27} = \frac{2}{3}$ --- **Angle $\alpha$ value:** $$\tan \alpha = \frac{3}{2}$$ $$\alpha = \arctan\left(\frac{3}{2}\right) \approx 56.30^\circ$$ --- **Summary:** - Hypotenuse $RQ = 9\sqrt{13}$ - $\csc \alpha = \frac{\sqrt{13}}{3}$ - $\sin \alpha = \frac{3\sqrt{13}}{13}$ - $\tan \alpha = \frac{3}{2}$ - $\cos \alpha = \frac{2\sqrt{13}}{13}$ - $\cot \alpha = \frac{2}{3}$ - $\sec \alpha = \frac{\sqrt{13}}{2}$ - Angle $\alpha \approx 56.30^\circ$