1. We are given a right triangle XYZ with right angle at O on the base XY. 2. We know: - Base XY = 34 cm - Height OZ = 16 cm (perpendicular to XY) - Angle OZY = 63°, the angle between OZ and ZY 3. We want to find the length of side XZ. 4. Since OZ is perpendicular to XY at point O, triangle XOZ is a right triangle with right angle at O. 5. To find XZ, note that point O lies on XY. Let XO = $x$ and OY = $34 - x$ because XY = 34. 6. Consider triangle OZY. Angle OZY = 63° and OZ = 16 cm. ZY is the hypotenuse of triangle OZY. 7. Using sine definition in triangle OZY: $$\sin(63^\circ) = \frac{OZ}{ZY} = \frac{16}{ZY} \implies ZY = \frac{16}{\sin(63^\circ)}$$ 8. Calculate $ZY$: $$ZY = \frac{16}{\sin(63^\circ)} \approx \frac{16}{0.8910} \approx 17.95 \text{ cm}$$ 9. Use cosine definition to find OY in triangle OZY: $$\cos(63^\circ) = \frac{OY}{ZY} \implies OY = ZY \times \cos(63^\circ)$$ 10. Calculate $OY$: $$OY = 17.95 \times 0.4540 \approx 8.15 \text{ cm}$$ 11. Since XY = 34 cm and OY = 8.15 cm, then: $$XO = XY - OY = 34 - 8.15 = 25.85 \text{ cm}$$ 12. Finally, use Pythagoras theorem in triangle XOZ (right triangle at O): $$XZ = \sqrt{XO^2 + OZ^2} = \sqrt{25.85^2 + 16^2} = \sqrt{668.22 + 256} = \sqrt{924.22}$$ 13. Calculate XZ: $$XZ \approx 30.4 \text{ cm}$$ **Answer:** The length of side XZ is approximately 30.4 cm, to 1 decimal place.