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Maths · Trigonometrical identities and trigonometrical functions

The angle of elevation of the top of a vertical tower from a point due east of i

The angle of elevation of the top of a vertical tower from a point \( A, \) due east of it is \( 45^{\circ} . \) The angle of elevation of the top of the same tower from a point \( \boldsymbol{B} \) due south of \( A \) is \( 30^{\circ} . \) If the distance between \( A \) and \( B \) is \( 54 \sqrt{2} m, \) then the height of the tower (in metres), is :

  • A. \( 36 \sqrt{3} \)
  • B. 108
  • C. 54 \)
  • D. \( 54 \sqrt{3} \)

Step-by-step solution

Let h be the height of the tower. From point A (due east), tan45° = h/OA => OA = h. From point B (due south of A), tan30° = h/OB => OB = h√3. The ground triangle OAB is right-angled at A (OA east, AB south). So by Pythagoras: OB² = OA² + AB² => (h√3)² = h² + (54√2)² => 3h² = h² + 54² * 2 => 2h² = 54² * 2 => h² = 54² => h = 54 m.
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