**Step 1: Introduction to the Concept** This problem involves applying geometric principles (law of sines, triangle area formulas), and concepts of circles (sector area and arc length) to solve real-world questions about a soap box derby car's design. We'll break down each question and solve it step by step. --- **Step 2:** ### 1. Find the length of the side opposite the obtuse angle (130°) We'll use the Law of Sines: \[ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} \] Given: Angle A = 130° (obtuse), angle B = 19°, and the side between them is 7 ft. First, find the third angle: \[ C = 180° - 130° - 19° = 31° \] Let side \( a \) be opposite 130°, side \( b \) opposite 19°, and side \( c = 7 \) ft opposite 31°. \[ \frac{a}{\sin 130°} = \frac{7}{\sin 31°} \] \[ a = \frac{7 \cdot \sin 130°}{\sin 31°} \] \[ a \approx \frac{7 \cdot 0.7660}{0.5150} \approx \frac{5.362}{0.5150} \approx 10.42 \text{ ft} \] **Answer:** \( a \approx 10.42 \) ft --- **Step 3:** ### 2. Area of the side panel (triangle) Use the formula for the area of a triangle given two sides and the included angle: \[ \text{Area} = \frac{1}{2} ab \sin C \] We know sides \( 7 \) ft and \( 10.42 \) ft with included angle \( 19° \): \[ \text{Area} = \frac{1}{2} \times 7 \times 10.42 \times \sin 19° \] \[ = 36.47 \times 0.3256 \approx 11.88 \text{ ft}^2 \] **Answer:** \( \text{Area} \approx 11.88 \text{ ft}^2 \) --- **Step 4:** ### 3. Area of the bright yellow triangle (sides: 2 ft, 3.5 ft, 5 ft) Use Heron's formula: \[ s = \frac{2 + 3.5 + 5}{2} = 5.25 \] \[ \text{Area} = \sqrt{s(s-a)(s-b)(s-c)} \] \[ = \sqrt{5.25(5.25-2)(5.25-3.5)(5.25-5)} \] \[ = \sqrt{5.25 \cdot 3.25 \cdot 1.75 \cdot 0.25} \] \[ = \sqrt{5.25 \cdot 3.25 \cdot 0.4375} \] \[ = \sqrt{5.25 \cdot 1.421875} \] \[ = \sqrt{7.460} \] \[ \approx 2.73 \text{ ft}^2 \] **Answer:** \( \text{Area} \approx 2.73 \text{ ft}^2 \) --- **Step 5:** ### 4. Area painted red on one front wheel Total area = \(\pi r^2 = \pi \times 8^2 = 64\pi\) in\(^2\) Let the green sector = \(x\). Blue = \(2x\), Red = \(2x\). Total: \[ x + 2x + 2x = 5x = 64\pi \implies x = \frac{64\pi}{5} \] Red area = \(2x = 2 \times \frac{64\pi}{5} = \frac{128\pi}{5}\) \[ \approx \frac{128 \times 3.1416}{5} \approx \frac{402.12}{5} \approx 80.42 \text{ in}^2 \] **Answer:** \( \approx 80.42 \) in\(^2\) --- **Step 6:** ### 5. Radius of rear wheel (fraction) Let \( r \) = radius of rear wheel, \( 8 \) = radius of front wheel. The arc length for full turn front = \(2\pi \times 8 = 16\pi\) Arc length for \( \frac{4}{5} \) turn rear = \(2\pi r \times \frac{4}{5}\) Set equal: \[ 16\pi = 2\pi r \times \frac{4}{5} \] \[ 16\pi = \frac{8\pi r}{5} \] \[ 16\pi \times 5 = 8\pi r \implies 80\pi = 8\pi r \implies r = 10 \] **Answer:** \( r = 10 \) ft --- **Step 7:** ### 6. Central angle for arc length (radius 12 in) Arc length needed = full turn of 10 ft wheel = \(2\pi \times 10 = 20\pi\) ft \(= 240\pi\) in. Arc length \( = r\theta \), so: \[ 12\theta = 240\pi \implies \theta = \frac{240\pi}{12} = 20\pi \text{ radians} \] **Answer:** \( \theta = 20\pi \) radians --- **Final Step: Summary in a Few Sentences and Step Explanation** **Whole Final Answer:** 1. The side opposite the obtuse angle is about 10.42 ft. 2. The area of the side panel is about 11.88 ft². 3. The yellow triangle's area is about 2.73 ft². 4. The red area on one front wheel is about 80.42 in². 5. The rear wheel radius is 10 ft. 6. The central angle needed for the 12 in rear wheel is \(20\pi\) radians. **What we did in each step:** Step 1: Introduced the geometric and circle concepts involved; Step 2: Used the Law of Sines; Step 3: Used the triangle area formula; Step 4: Used Heron's formula; Step 5: Used circle area logic; Step 6: Used proportional arc lengths; Step 7: Used arc length formula.