1. **State the problem:** We have a semicircle with center O and radius 25 cm. M is the midpoint of chord AB, and OM = 18 cm. We need to (a) show that angle AOB = 87.9° correct to 1 decimal place. 2. **Recall the cosine rule:** $$\cos A = \frac{b^2 + c^2 - a^2}{2bc}$$ where $A$ is the angle opposite side $a$, and $b$, $c$ are the other two sides. 3. **Identify the triangle and sides:** In triangle AOB, OA = OB = 25 cm (radii), and AB is the chord. We want to find angle AOB. 4. **Calculate length AB:** Since M is midpoint of AB and OM = 18 cm, and O is center, OM is perpendicular to AB, so triangle OMB is right angled. Using Pythagoras theorem in triangle OMB: $$MB = \sqrt{OB^2 - OM^2} = \sqrt{25^2 - 18^2} = \sqrt{625 - 324} = \sqrt{301} \approx 17.35 \text{ cm}$$ Since M is midpoint, $$AB = 2 \times MB = 2 \times 17.35 = 34.7 \text{ cm}$$ 5. **Apply cosine rule to triangle AOB:** Sides: $a = AB = 34.7$, $b = OA = 25$, $c = OB = 25$ Calculate: $$\cos \angle AOB = \frac{b^2 + c^2 - a^2}{2bc} = \frac{25^2 + 25^2 - 34.7^2}{2 \times 25 \times 25} = \frac{625 + 625 - 1204.09}{1250} = \frac{45.91}{1250} = 0.0367$$ 6. **Find angle AOB:** $$\angle AOB = \cos^{-1}(0.0367) \approx 87.9^\circ$$ **Final answer:** $$\boxed{87.9^\circ}$$