1. **Problem statement:** ABCD is a rectangle with length 16 m and width 6 m. Find angles $a$ and $b$ to the nearest degree. 2. **Recall properties of rectangles:** All angles are right angles ($90^\circ$). The diagonals are equal in length and bisect each other. 3. **Calculate the diagonal length using the Pythagorean theorem:** $$d = \sqrt{16^2 + 6^2} = \sqrt{256 + 36} = \sqrt{292}$$ 4. **Calculate $a$ (angle between diagonal and side 16 m):** Use cosine rule: $$\cos a = \frac{\text{adjacent side}}{\text{hypotenuse}} = \frac{16}{\sqrt{292}}$$ 5. **Calculate $a$:** $$a = \cos^{-1}\left(\frac{16}{\sqrt{292}}\right) = \cos^{-1}(0.936) \approx 20^\circ$$ 6. **Calculate $b$ (angle between diagonals at center):** Diagonals of a rectangle bisect each other and form two congruent triangles. The angle between diagonals is twice $a$ because diagonals intersect symmetrically. 7. **Calculate $b$:** $$b = 2a = 2 \times 20^\circ = 40^\circ$$ **Final answers:** $$a \approx 20^\circ, \quad b \approx 40^\circ$$