1. **State the problem:** An aeroplane flying at 6000 m height passes vertically above another plane, and from an observer's point on the ground, the angles of elevation to the two planes are 60° and 45° respectively. 2. **Identify what is asked:** Find how much higher the upper plane is than the lower plane. 3. **Set variables:** Let the horizontal distance from the observer to the planes' vertical line be $x$ meters. Let the height of the lower plane be $h$ meters. 4. **Use the angles with tangent:** From the observer, the angle of elevation to the top plane is 60°, so: $$\tan 60^\circ = \frac{6000}{x}$$ But $\tan 60^\circ = \sqrt{3}$, so $$\sqrt{3} = \frac{6000}{x} \implies x = \frac{6000}{\sqrt{3}}$$ 5. **Use the angle of elevation to the lower plane (45°):** $$\tan 45^\circ = \frac{h}{x}$$ But $\tan 45^\circ = 1$, thus $$1 = \frac{h}{x} \implies h = x$$ 6. **Calculate $h$ using $x$: ** $$h = \frac{6000}{\sqrt{3}}$$ 7. **Calculate how much higher the top plane is:** $$6000 - h = 6000 - \frac{6000}{\sqrt{3}} = 6000 \left(1 - \frac{1}{\sqrt{3}}\right)$$ 8. **Simplify the difference:** $$= 6000 \left(\frac{\sqrt{3} - 1}{\sqrt{3}}\right) = \frac{6000(\sqrt{3} - 1)}{\sqrt{3}}$$ **Final answer:** The top plane is $\frac{6000(\sqrt{3} - 1)}{\sqrt{3}}$ meters higher than the lower plane, approximately $1780$ meters higher.