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      Question

      From a point on level ground, the angle of elevation of

      the top of a vertical tower is 30Β°. On moving 30 m closer to the tower, the angle of elevation becomes 60Β°. Find the height of the tower.
      A 13√3 m Correct Answer Incorrect Answer
      B 17√3 m Correct Answer Incorrect Answer
      C 15√3 m Correct Answer Incorrect Answer
      D 25√3 m Correct Answer Incorrect Answer

      Solution

      Let the initial horizontal distance be x m and height of tower be h m. From first position: tan 30Β° = h/x β‡’ 1/√3 = h/x β‡’ h = x/√3 From second position: distance = x βˆ’ 30, angle = 60Β°: tan 60Β° = h/(x βˆ’ 30) β‡’ √3 = h/(x βˆ’ 30) β‡’ h = √3(x βˆ’ 30) Equate h: x/√3 = √3(x βˆ’ 30) Multiply by √3: x = 3(x βˆ’ 30) x = 3x βˆ’ 90 2x = 90 β‡’ x = 45 m Then h = x/√3 = 45/√3 = 15√3 m.

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