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    Question

    A man observes the top of a tower at an angle of

    elevation of 30Β°. He walks 100 meters closer to the tower, and the angle of elevation becomes 60Β°. If the height of the tower remains constant, find the initial distance of the man from the base of the tower.
    A 100 m Correct Answer Incorrect Answer
    B 150 m Correct Answer Incorrect Answer
    C 150√3 m Correct Answer Incorrect Answer
    D 200√3 m Correct Answer Incorrect Answer

    Solution

    Let the height of the tower be h meters, and the initial distance from the man to the base of the tower be x meters. From the first position: tan 30Β° = h / x β†’ 1/√3 = h / xΒ  x = h√3. From the second position: tan 60Β° = h / (x - 100)Β  √3 = h / (x - 100). Substitute x = h√3: √3 = h / (h√3 - 100). Cross-multiply: h√3 - 100 = h / √3. 3h - 100√3 = h. 2h = 100√3. h = 50√3. Now, x = h√3 = 50√3 Γ— √3 = 150 m. Answer: b) 150 m

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