1. **State the problem:** Find the angle $\angle XYZ$ in the given right triangle with vertices $X$, $Y$, and $Z$. 2. **Identify the sides:** From the description, the triangle $XYZ$ has sides: - $XV = 17$ cm - $VZ = 8$ cm - $YZ = 10$ cm Since $V$ is the foot of the perpendicular from $Y$ to $XZ$, triangle $XYZ$ is right angled at $V$. 3. **Use the Pythagorean theorem:** To confirm the right triangle, check if $XV^2 + VZ^2 = YZ^2$: $$17^2 + 8^2 = 289 + 64 = 353$$ $$10^2 = 100$$ Since $353 \neq 100$, the right angle is at $V$, but $YZ$ is not the hypotenuse of triangle $XVZ$. The triangle $XYZ$ is composed of two right triangles sharing $V$. 4. **Find $XY$ using Pythagoras in triangle $XYV$:** Since $YV$ is perpendicular to $XZ$, and $XV = 17$ cm, $VZ = 8$ cm, and $YZ = 10$ cm, we can find $XY$ by: $$XY = \sqrt{XV^2 + YV^2}$$ But $YV$ is unknown. However, since $YZ = 10$ cm and $VZ = 8$ cm, $YV$ can be found by: $$YV = \sqrt{YZ^2 - VZ^2} = \sqrt{10^2 - 8^2} = \sqrt{100 - 64} = \sqrt{36} = 6$$ 5. **Calculate $XY$:** $$XY = \sqrt{XV^2 + YV^2} = \sqrt{17^2 + 6^2} = \sqrt{289 + 36} = \sqrt{325}$$ 6. **Find angle $\angle XYZ$:** This angle is at vertex $Y$, between sides $XY$ and $YZ$. Use the cosine rule: $$\cos(\angle XYZ) = \frac{XY^2 + YZ^2 - XZ^2}{2 \cdot XY \cdot YZ}$$ But $XZ = XV + VZ = 17 + 8 = 25$ cm. Calculate: $$\cos(\angle XYZ) = \frac{325 + 100 - 625}{2 \cdot \sqrt{325} \cdot 10} = \frac{-200}{20 \sqrt{325}} = \frac{-10}{\sqrt{325}}$$ 7. **Calculate the angle:** $$\angle XYZ = \cos^{-1}\left(\frac{-10}{\sqrt{325}}\right)$$ Calculate numeric value: $$\sqrt{325} \approx 18.03$$ $$\frac{-10}{18.03} \approx -0.555$$ So, $$\angle XYZ \approx \cos^{-1}(-0.555) \approx 123.7^\circ$$ **Final answer:** $$\boxed{\angle XYZ \approx 123.7^\circ}$$