1. **Problem:** A balloon is 2500 ft above a lake. The angles of depression to the two opposite shores are 43° and 27°. We need to find the width of the lake. 2. **Setup:** Let the balloon be at point B, exactly 2500 ft above the lake surface. 3. Let points A and C be the two points on opposite shores on the lake's surface. We want to find the distance $AC$ (the width of the lake). 4. Draw horizontal line from B parallel to the lake surface, and drop vertical 2500 ft from B to the lake (point O). 5. From the balloon, the angle of depression to point A is 43°, so angle $BAO = 43^\circ$ (alternate interior angles). 6. Similarly, angle of depression to point C is 27°, so angle $BCO = 27^\circ$. 7. In right triangle $ABO$, tan 43° = $\frac{BO}{AO} = \frac{2500}{AO}$, so $$AO = \frac{2500}{\tan 43^\circ}$$ 8. In right triangle $BCO$, tan 27° = $\frac{BO}{OC} = \frac{2500}{OC}$, so $$OC = \frac{2500}{\tan 27^\circ}$$ 9. The width of the lake $AC = AO + OC$ because both points are on opposite sides from directly beneath the balloon. 10. Calculate using $\tan 43^\circ \approx 0.9325$ and $\tan 27^\circ \approx 0.5095$: $$AO = \frac{2500}{0.9325} \approx 2681.11$$ $$OC = \frac{2500}{0.5095} \approx 4907.40$$ 11. So, the width $AC = 2681.11 + 4907.40 = 7588.51$ ft. **Final answer:** The width of the lake is approximately $7588.51$ ft.