A man 1.6 m tall stands 25 m from a cell tower. The angle of elevation of the top of the tower from his eyes is 60°. Find the height of the tower. (√3 = 1.732)

1. 1 · Write what is given and what is to be found Given: eye height 1.6 m, distance 25 m, angle of elevation 60°. To find: the height of the tower.
2. 2 · Draw the figure Tower AB (A top, B foot). Man CD, with his eye at D. Draw the horizontal line DE to meet the tower at E, so DE = 25 m and EB = 1.6 m. Mark ∠ADE = 60° and the right angle at E.
3. 3 · Write the formula tan θ = opposite/adjacent, and tan 60° = √3.
4. 4 · Find the height above eye level
   • In right triangle AED, tan 60° = AE/DE
   • √3 = AE/25
   • AE = 25√3 = 25 × 1.732 = 43.30 m
5. 5 · Add the man's height AB = AE + EB = 43.30 + 1.6 = 44.90 m
6. 6 · Does it make sense? Cell towers are often a few tens of metres tall, so about 45 m is a sensible answer.
7. 7 · Check and box the answer Height of the tower = 44.90 m

Scoring tip: Do not forget to add the observer's height at the end. It is the most common lost mark in this type of question.