1. **State the problem:** Jack and Jill are on opposite sides of an 88 metre deep canyon. Jack sees the trail guide at an angle of depression of 45° and Jill sees the trail guide at an angle of depression of 60°. We need to find the distance between Jack and Jill. 2. **Set up the scenario:** Let the canyon depth be the vertical height $h = 88$ metres. 3. **Use trigonometry:** The angles of depression correspond to angles between the horizontal line at Jack's and Jill's positions and the line of sight to the trail guide. 4. **Define distances:** Let $x$ be the horizontal distance from Jack to the trail guide, and $y$ be the horizontal distance from Jill to the trail guide. 5. **Apply tangent function:** For Jack, $$\tan(45^\circ) = \frac{h}{x}$$ Since $\tan(45^\circ) = 1$, we have $$1 = \frac{88}{x} \implies x = 88$$ 6. For Jill, $$\tan(60^\circ) = \frac{h}{y}$$ Since $\tan(60^\circ) = \sqrt{3}$, we have $$\sqrt{3} = \frac{88}{y} \implies y = \frac{88}{\sqrt{3}}$$ 7. **Simplify Jill's distance:** Rationalize the denominator: $$y = \frac{88}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{88\sqrt{3}}{3}$$ 8. **Find total distance between Jack and Jill:** $$d = x + y = 88 + \frac{88\sqrt{3}}{3} = 88\left(1 + \frac{\sqrt{3}}{3}\right)$$ 9. **Approximate the value:** $$\sqrt{3} \approx 1.732$$ $$d \approx 88 \times \left(1 + \frac{1.732}{3}\right) = 88 \times (1 + 0.577) = 88 \times 1.577 = 138.776$$ **Final answer:** The distance between Jack and Jill is approximately **138.78 metres**.