Mathematics

Mathematics

Problem

A vertical tower stands on level ground. From point A the angle of elevation of the top of the tower is θ\theta. After walking 40 m40\,\text{m} straight towards the tower to point B, the angle of elevation becomes (90∘−θ)(90^\circ-\theta). a. (i) Draw a neat labelled diagram to represent the above situation. b. (ii) Using the diagram, express the height hh of the tower in terms of θ\theta. c. (iii) Calculate hh when θ=28∘\theta = 28^\circ. (Take sin⁡28∘=0.4695\sin 28^\circ = 0.4695 and cos⁡28∘=0.8829\cos 28^\circ = 0.8829, correct to four decimal places.)
Intuitive explanation

The key insight is recognizing that the two angles of elevation are complementary — they sum to 90°90° — which creates a special geometric relationship. When you set up right triangles from both observation points, the complementary angles mean the two triangles are similar in a useful way: the ratio of distances and heights will satisfy a constraint that lets you solve for the tower's height. Use the tangent ratio from each position and the fact that the horizontal distance decreases by 40 m40\,\text{m} to build your equations.

Intuitive explanation

The key insight is recognizing that the two angles of elevation are complementary — they sum to 90°90° — which creates a special geometric relationship. When you set up right triangles from both observation points, the complementary angles mean the two triangles are similar in a useful way: the ratio of distances and heights will satisfy a constraint that lets you solve for the tower's height. Use the tangent ratio from each position and the fact that the horizontal distance decreases by 40 m40\,\text{m} to build your equations.

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