1. **Problem Statement:** Find the height of the building given a right triangle formed by the building, a car, and an observer. The angles from the base are 60° and 30°, and the height is labeled as $h$. The distance from the observer to the car and the building is related to these angles. 2. **Given:** - Angle at observer to car: 60° - Angle at observer to building: 30° - Height of building: $h$ - Distance from observer to car: $x$ - Distance from car to building: $2S\sqrt{3}$ (given) 3. **Formula and Rules:** - Use the tangent function in right triangles: $\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}}$ - For angle 60°, $\tan 60° = \sqrt{3}$ - For angle 30°, $\tan 30° = \frac{1}{\sqrt{3}}$ 4. **Step 1: Express height $h$ in terms of $x$ using angle 60°** $$h = x \tan 60° = x \sqrt{3}$$ 5. **Step 2: Express height $h$ in terms of $x$ and distance from car to building using angle 30°** The total distance from observer to building is $x + 2S\sqrt{3}$. $$h = (x + 2S\sqrt{3}) \tan 30° = (x + 2S\sqrt{3}) \frac{1}{\sqrt{3}}$$ 6. **Step 3: Equate the two expressions for $h$** $$x \sqrt{3} = \frac{x + 2S\sqrt{3}}{\sqrt{3}}$$ 7. **Step 4: Multiply both sides by $\sqrt{3}$ to clear denominator** $$3x = x + 2S\sqrt{3}$$ 8. **Step 5: Solve for $x$** $$3x - x = 2S\sqrt{3}$$ $$2x = 2S\sqrt{3}$$ $$x = S\sqrt{3}$$ 9. **Step 6: Find height $h$ using $x$** $$h = x \sqrt{3} = S\sqrt{3} \times \sqrt{3} = 3S$$ **Final answer:** The height of the building is $$h = 3S$$. --- **Slug:** building height **Subject:** trigonometry **Desmos:** {"latex":"y=3S","features":{"intercepts":true,"extrema":true}} **q_count:** 1