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5-mark answer frame

The angle of elevation of the top of a tower from a point on the ground is 30°. After walking 50 m towards the tower, the angle becomes 60°. Find the height of the tower. (Use √3 = 1.732)

How to lay out the answer. Follow the frame, then write it in your own words.

  1. 1 · Write what is given and what is to be found Given: tower AB = h; points C and D on the ground with DC = 50 m; ∠ADB = 30°, ∠ACB = 60°. To find: h.
  2. 2 · Draw the figure Vertical AB with foot B. C (nearer) and D (farther) on line DB. Join AC and AD. Mark 60° at C and 30° at D.
  3. 3 · Write the formula or theorem tan θ = opposite/adjacent; tan 60° = √3, tan 30° = 1/√3.
  4. 4 · Equation from ΔABC tan 60° = h/x, so h = √3 x ... (1)
  5. 5 · Equation from ΔABD tan 30° = h/(x + 50), so √3 h = x + 50 ... (2)
  6. 6 · Solve Put (1) in (2): 3x = x + 50, so x = 25 m. Then h = 25√3 = 25 × 1.732 = 43.30 m.
  7. 7 · Check and box the answer The tower is 43.3 m high. Check: 43.3 ÷ 25 = 1.732 = tan 60°.

Scoring tip: Number your two equations (1) and (2) and write 'substituting (1) in (2)'. That one line shows the examiner your method clearly.