A person initially observes the top of a building at an angle of elevation of 45 degrees. After walking 100 metres straight towards the building along level ground, the angle of elevation becomes 60 degrees. What is the height of the building in metres?

Introduction / Context:
This problem tests understanding of how angles of elevation change as an observer moves closer to a vertical object. The observer walks towards a building and the angle of elevation of the building top increases from 45 degrees to 60 degrees. Using trigonometry, and particularly the tangent function in two right angled triangles that share the same height, we can determine the unknown height of the building.

Given Data / Assumptions:

• Initial angle of elevation of the top of the building is 45 degrees.
• After walking 100 metres towards the building, the angle of elevation becomes 60 degrees.
• The observer walks along level ground in a straight line directly towards the building.
• The building is vertical and stands on the same horizontal level as the observer.
• Let the height of the building be h metres.

Concept / Approach:
For a right angled triangle, tan of the angle of elevation equals the ratio of the height of the object to the horizontal distance between the observer and the object. If the initial horizontal distance is x metres, then tan 45 degrees gives h in terms of x, and tan 60 degrees gives h in terms of x − 100, because the observer has walked 100 metres closer. Solving this system of equations allows us to find x and then h in exact form.

Step-by-Step Solution:
Step 1: Let the initial distance from the building be x metres. With angle 45°, tan 45° = h / x, so 1 = h / x, which implies h = x. Step 2: After walking 100 metres towards the building, the new distance is x − 100 metres. Step 3: With angle 60°, tan 60° = h / (x − 100). Since tan 60° = √3 and h = x, we get √3 = x / (x − 100). Step 4: Cross multiply to obtain √3(x − 100) = x, which simplifies to √3 x − 100√3 = x. Step 5: Rearrange to get (√3 − 1)x = 100√3, so x = 100√3 / (√3 − 1). Step 6: Rationalise the denominator: x = 100√3(√3 + 1) / (3 − 1) = 50(3 + √3). Step 7: Since h = x, the height of the building is h = 50(3 + √3) metres.

Verification / Alternative check:
Approximate √3 as 1.732. Then 3 + √3 is about 4.732 and h ≈ 50 × 4.732 ≈ 236.6 metres. Initially, distance x is also 236.6 metres, giving tan 45° = 236.6 / 236.6 = 1, correct. After moving 100 metres, the new distance is about 136.6 metres, and tan 60° ≈ 236.6 / 136.6 ≈ 1.732, which matches √3 and confirms the correctness of the answer.

Why Other Options Are Wrong:
Option 100(√3 + 1) corresponds to a different height that does not satisfy both angle conditions. Option 150 and option 100√3 produce approximate distances that fail to give tan 45° and tan 60° simultaneously. Option 200 is a simple approximate value and does not satisfy the trigonometric equations derived from the two angles.

Common Pitfalls:
Learners sometimes forget that the second horizontal distance is x minus 100, not x plus 100. Another common error is to use sine instead of tangent, or to incorrectly rationalise the denominator when simplifying x. Some students also approximate √3 too early, which can lead to rounding errors and incorrect identification of the exact form of the height.

Final Answer:
The height of the building is 50(3 + √3) metres.