1. **Problem Statement:** An airplane A is flying at an elevation of 5625 ft above a straight highway. Two cars P and Q are on opposite sides of the plane on the highway. The angle of elevation from car P to the plane is 38° and from car Q to the plane is 46°. 2. **Diagram Explanation:** - Point A is directly above the highway at height 5625 ft. - Points P and Q lie on the highway on opposite sides of the vertical line from A. - The angles of elevation from P and Q to A are 38° and 46° respectively. 3. **Goal:** Find the distance between cars P and Q. 4. **Formula and Approach:** We use right triangle trigonometry. Let the horizontal distances from P to the point under A be $x$ and from Q to the point under A be $y$. From the right triangles formed: $$\tan(38^\circ) = \frac{5625}{x} \implies x = \frac{5625}{\tan(38^\circ)}$$ $$\tan(46^\circ) = \frac{5625}{y} \implies y = \frac{5625}{\tan(46^\circ)}$$ The total distance between cars P and Q is: $$d = x + y$$ 5. **Calculations:** Calculate $x$: $$x = \frac{5625}{\tan(38^\circ)}$$ Using $\tan(38^\circ) \approx 0.7813$: $$x \approx \frac{5625}{0.7813} \approx 7200.1 \text{ ft}$$ Calculate $y$: $$y = \frac{5625}{\tan(46^\circ)}$$ Using $\tan(46^\circ) \approx 1.0355$: $$y \approx \frac{5625}{1.0355} \approx 5431.3 \text{ ft}$$ 6. **Final answer:** $$d = x + y \approx 7200.1 + 5431.3 = 12631.4 \text{ ft}$$ **Therefore, the cars are approximately 12631.4 feet apart.**