1. Stating the problem: We have two airplanes observed from the same point on the ground. The higher plane is 6000 m above the ground. The angles of elevation to the two planes are 60° and 45°, respectively. We need to find how much higher the top plane is than the lower plane. 2. Let the distance from the observer to the point on the ground vertically below both planes be $x$ meters. The height of the lower plane is $h$ meters, and the height of the top plane is 6000 m. 3. Using the angle of elevation 60° for the top plane, we write: $$ \tan 60^\circ = \frac{6000}{x} $$ We know $\tan 60^\circ = \sqrt{3}$. So, $$ \sqrt{3} = \frac{6000}{x} \implies x = \frac{6000}{\sqrt{3}} $$ 4. Using the angle of elevation 45° for the lower plane: $$ \tan 45^\circ = \frac{h}{x} $$ Since $\tan 45^\circ = 1$, we have: $$ 1 = \frac{h}{x} \implies h = x $$ 5. Substitute $x$ from step 3 into $h$: $$ h = \frac{6000}{\sqrt{3}} $$ 6. The difference in height between the two planes is: $$ 6000 - h = 6000 - \frac{6000}{\sqrt{3}} = 6000 \left(1 - \frac{1}{\sqrt{3}} \right) $$ 7. Simplify the expression inside the parentheses: $$ 1 - \frac{1}{\sqrt{3}} = \frac{\sqrt{3} - 1}{\sqrt{3}} $$ Thus, $$ \text{Height difference} = 6000 \times \frac{\sqrt{3} - 1}{\sqrt{3}} $$ 8. Approximating numerically: $$ \sqrt{3} \approx 1.732 $$ $$ \text{Height difference} \approx 6000 \times \frac{1.732 - 1}{1.732} = 6000 \times \frac{0.732}{1.732} \approx 6000 \times 0.4226 = 2536 $$ Final answer: The plane on top is approximately 2536 metres higher than the other plane.