1. **Problem (a): Find the size of \(\angle BAC\) to 1 decimal place.** 2. To find \(\angle BAC\), we need the coordinates or lengths of sides of triangle \(ABC\) or additional information, which is not provided here. Please provide the necessary data to proceed. 3. **Problem (b): Find the exact value of the area of \(\triangle ABC\).** 4. Similarly, to find the area of \(\triangle ABC\), we need side lengths or height or coordinates. Please provide these details. --- 5. **Problem 2: School running track area and maximum area.** 6. The track consists of two straight sections (length \(l\)) and two semi-circular ends each with radius \(r\). 7. The total length of the track is given as \(300\) m, so the perimeter is: $$ 2l + 2\pi r = 300 $$ 8. Solving for \(l\): $$ l = \frac{300 - 2\pi r}{2} = 150 - \pi r $$ 9. The internal area \(A\) consists of the rectangular part plus the area of the two semi-circles (which make a full circle): $$ A = l \times 2r + \pi r^2 $$ 10. Substitute \(l = 150 - \pi r\): $$ A = (150 - \pi r) \times 2r + \pi r^2 = 300r - 2\pi r^2 + \pi r^2 = 300r - \pi r^2 $$ This shows the formula: $$ A = 300r - \pi r^2 $$ 11. To find the maximum value of \(A\), differentiate with respect to \(r\) and set to zero: $$ \frac{dA}{dr} = 300 - 2\pi r = 0 $$ 12. Solve for \(r\): $$ 2\pi r = 300 \implies r = \frac{300}{2\pi} = \frac{150}{\pi} $$ 13. Substitute \(r = \frac{150}{\pi}\) into \(A\): $$ A = 300 \times \frac{150}{\pi} - \pi \left(\frac{150}{\pi}\right)^2 = \frac{45000}{\pi} - \pi \times \frac{22500}{\pi^2} = \frac{45000}{\pi} - \frac{22500}{\pi} = \frac{22500}{\pi} $$ 14. **Final answers:** - (a) and (b) require more data. - Area formula: \( A = 300r - \pi r^2 \) - Maximum area: \( \frac{22500}{\pi} \) square meters.