1. **Problem statement:** A surveyor in an airplane observes the angle of depression to the near side of a lake as 45° and to the far side as 32°. The airplane is 9750 m from the near side. We need to find the width of the lake. 2. **Understanding the problem:** The airplane, near side, and far side form a right triangle setup with angles of depression given. The width of the lake is the horizontal distance between the near and far sides. 3. **Key formula:** Using trigonometry, specifically the tangent of the angles of depression: $$\tan(\theta) = \frac{\text{height}}{\text{horizontal distance}}$$ 4. **Assign variables:** - Let $h$ be the height of the airplane above the lake. - Distance from airplane to near side horizontally is 9750 m. - Angle of depression to near side: $45^\circ$. - Angle of depression to far side: $32^\circ$. - Let $d$ be the width of the lake (distance between near and far sides). 5. **Calculate height $h$ using near side:** Since angle of depression is $45^\circ$, $$\tan 45^\circ = \frac{h}{9750}$$ We know $\tan 45^\circ = 1$, so $$1 = \frac{h}{9750} \implies h = 9750 \text{ m}$$ 6. **Calculate horizontal distance to far side:** Let $x$ be the horizontal distance from airplane to far side. Using angle $32^\circ$: $$\tan 32^\circ = \frac{h}{x} = \frac{9750}{x}$$ Rearranged: $$x = \frac{9750}{\tan 32^\circ}$$ Calculate $\tan 32^\circ \approx 0.6249$, $$x \approx \frac{9750}{0.6249} \approx 15606.7 \text{ m}$$ 7. **Find width of lake $d$:** $$d = x - 9750 = 15606.7 - 9750 = 5856.7 \text{ m}$$ **Final answer:** The width of the lake is approximately **5857 meters**.