πŸ“’ Too many exams? Don’t know which one suits you best? Book Your Free Expert πŸ‘‰ call Now!

    Question

    The angle of elevation of the top of a tower from a

    point on the ground is 30Β°. On moving 20 m closer to the tower, the angle of elevation becomes 45Β°. Find the height of the tower.
    A 13(√3 + 2) m. Correct Answer Incorrect Answer
    B 10(√3 + 1) m Correct Answer Incorrect Answer
    C 11(√3 - 1) m. Correct Answer Incorrect Answer
    D 13(√2 + 1) m. Correct Answer Incorrect Answer

    Solution

    ATQ, Let height of tower = h, initial horizontal distance = x. From first position: tan 30Β° = h/x β‡’ 1/√3 = h/x β‡’ h = x/√3 From second position: distance = x βˆ’ 20, angle = 45Β°: tan 45Β° = h/(x βˆ’ 20) = 1 β‡’ h = x βˆ’ 20 Equate: x βˆ’ 20 = x/√3 x(1 βˆ’ 1/√3) = 20 x[(√3 βˆ’ 1)/√3] = 20 x = 20√3 / (√3 βˆ’ 1) Rationalize: x = 20√3(√3 + 1) / [(√3 βˆ’ 1)(√3 + 1)] = 20√3(√3 + 1) / (3 βˆ’ 1) = 10√3(√3 + 1) = 10(3 + √3) Height h = x βˆ’ 20 = 10(3 + √3) βˆ’ 20 But from h = x/√3: h = [10(3 + √3)] / √3 = 10(3 + √3)/√3 Γ— (√3/√3) = 10(3√3 + 3)/3 = 10(√3 + 1) So height of tower = 10(√3 + 1) m.

    Practice Next
    More Height and Distance Questions
    ask-question