From two points on level ground that are 1 km apart, the angles of elevation of an aeroplane are 45° and 60°. What is the height of the aeroplane above the ground (in km)?

Introduction / Context:
This is a height and distance problem from trigonometry, often appearing in aptitude tests. It involves angles of elevation observed from two points on level ground and requires the use of tangent ratios to determine the height of an aeroplane.

Given Data / Assumptions:

• Two observation points on level ground are 1 km apart.
• The angles of elevation of the aeroplane from these points are 45 degrees and 60 degrees.
• The aeroplane is vertically above some point on the line connecting the two observers.
• We assume both observers and the foot of the perpendicular from the aeroplane lie on a straight horizontal line.

Concept / Approach:
We use the definition of tangent in a right triangle: tan(theta) = opposite / adjacent, where opposite is the height and adjacent is horizontal distance. If we let the nearer point (with larger angle 60 degrees) be B and the farther point (with smaller angle 45 degrees) be A, and the foot of the perpendicular from the aeroplane to the ground be D, we can form two right triangles A D and B D with common height but different base lengths.

Step-by-Step Solution:
Step 1: Let the height of the aeroplane above ground be h km.Step 2: Let A be the farther point and B be the nearer point, and let AD = x km.Step 3: Since A and B are 1 km apart on the same line, BD = x - 1 km.Step 4: From point A, tan 45° = h / x. Since tan 45° = 1, we get h = x.Step 5: From point B, tan 60° = h / (x - 1). Since tan 60° = √3, we have √3 = h / (x - 1).Step 6: Substitute h = x into the second equation to get √3 = x / (x - 1).Step 7: Cross-multiply: √3 (x - 1) = x.Step 8: Expand: √3 x - √3 = x.Step 9: Bring x terms together: √3 x - x = √3.Step 10: Factor x: x(√3 - 1) = √3, so x = √3 / (√3 - 1).Step 11: Rationalise the denominator: √3 / (√3 - 1) * (√3 + 1) / (√3 + 1) = (3 + √3) / (3 - 1) = (3 + √3) / 2.Step 12: Recall h = x, so the height h = (3 + √3) / 2 km.

Verification / Alternative check:
We can numerically approximate the height. √3 is about 1.732, so (3 + √3) / 2 is roughly (3 + 1.732) / 2 = 4.732 / 2 ≈ 2.366 km. If we use this approximate height and recompute the tangents for the two triangles, the angles will be close to 45 degrees and 60 degrees, confirming the solution.

Why Other Options Are Wrong:

• √3 km: This is about 1.732 km, which would not satisfy the geometry with a 1 km separation between observers and the given angles.
• (√3 + 1) km: Approximately 2.732 km, which leads to inconsistent tangent ratios.
• 2 km: This is a rough guess and does not satisfy both tan 45° and tan 60° relationships simultaneously.

Common Pitfalls:
Students may misplace the 1 km distance, assuming the aeroplane is directly over one of the observers, or they may swap which angle belongs to which point, leading to incorrect equations. Another frequent error is algebraic manipulation when solving for x and rationalising the denominator.

Final Answer:
The height of the aeroplane above the ground is (3 + √3) / 2 km.