1. **Problem 1: Vector triangle calculations** (a) Find the size of \(\angle BAC\) in degrees. 1. Given vectors: \[ \vec{AB} = -3\mathbf{i} + 6\mathbf{j}, \quad \vec{AC} = 10\mathbf{i} - 2\mathbf{j} \] 2. Use the dot product formula to find the angle between \(\vec{AB}\) and \(\vec{AC}\): \[ \vec{AB} \cdot \vec{AC} = |\vec{AB}| |\vec{AC}| \cos \theta \] 3. Calculate the dot product: \[ (-3)(10) + (6)(-2) = -30 - 12 = -42 \] 4. Calculate magnitudes: \[ |\vec{AB}| = \sqrt{(-3)^2 + 6^2} = \sqrt{9 + 36} = \sqrt{45} = 3\sqrt{5} \] \[ |\vec{AC}| = \sqrt{10^2 + (-2)^2} = \sqrt{100 + 4} = \sqrt{104} = 2\sqrt{26} \] 5. Substitute into the cosine formula: \[ \cos \theta = \frac{-42}{3\sqrt{5} \times 2\sqrt{26}} = \frac{-42}{6 \sqrt{130}} = \frac{-7}{\sqrt{130}} \] 6. Calculate \(\theta\): \[ \theta = \cos^{-1} \left( \frac{-7}{\sqrt{130}} \right) \approx \cos^{-1}(-0.6139) \approx 127.9^\circ \] (b) Find the exact value of the area of \(\triangle ABC\). 1. Area formula using vectors: \[ \text{Area} = \frac{1}{2} |\vec{AB} \times \vec{AC}| \] 2. Calculate the 2D cross product magnitude: \[ |\vec{AB} \times \vec{AC}| = |(-3)(-2) - (6)(10)| = |6 - 60| = 54 \] 3. Area: \[ \frac{1}{2} \times 54 = 27 \] --- 2. **Problem 2: Running track area optimization** (a) Show that the internal area \(A\) is given by \(A = 300r - \pi r^2\). 1. Let the length of the straight sections be \(l\) and radius of semicircles be \(r\). 2. The total length of the track is perimeter: \[ 2l + 2 \times (\text{circumference of semicircle}) = 2l + 2 \times \pi r = 300 \] 3. Simplify: \[ 2l + 2\pi r = 300 \implies l + \pi r = 150 \implies l = 150 - \pi r \] 4. The internal area consists of rectangle plus two semicircles (which make a full circle): \[ A = l \times 2r + \pi r^2 = 2r(150 - \pi r) + \pi r^2 = 300r - 2\pi r^2 + \pi r^2 = 300r - \pi r^2 \] (b) Find the maximum value of \(A\) in terms of \(\pi\). 1. Express \(A\) as a function of \(r\): \[ A(r) = 300r - \pi r^2 \] 2. Differentiate \(A\) with respect to \(r\): \[ \frac{dA}{dr} = 300 - 2\pi r \] 3. Set derivative to zero to find critical point: \[ 300 - 2\pi r = 0 \implies 2\pi r = 300 \implies r = \frac{300}{2\pi} = \frac{150}{\pi} \] 4. Substitute \(r = \frac{150}{\pi}\) into \(A(r)\): \[ A_{max} = 300 \times \frac{150}{\pi} - \pi \left( \frac{150}{\pi} \right)^2 = \frac{45000}{\pi} - \pi \times \frac{22500}{\pi^2} = \frac{45000}{\pi} - \frac{22500}{\pi} = \frac{22500}{\pi} \] **Final answers:** - \(\angle BAC \approx 127.9^\circ\) - Area of \(\triangle ABC = 27\) - Internal area formula: \(A = 300r - \pi r^2\) - Maximum internal area: \(\frac{22500}{\pi}\)