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A pole of height 5 m is fixed on the top of a tower. The angle of elevation of the top of the pole as observed from a point A on the ground is 60 and the angle of depression of the point A from the top of the tower is 45. Find the height of the tower. (Take √3 = 1.732)

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sir this is the important  question from the book -ML aggarwal( avichal publication) class 10th , chapter20 , heights and distances
this question has been asked in previous year exams
it is given that A pole of height 5 m is fixed on the top of a tower. The angle of elevation of the top of the pole as observed from a point A on the ground is 60 and the angle of depression of the point A from the top of the tower is 45. Find the height of the tower. (Take √3 = 1.732)

question no 33 , heights and distances , ICSE board, ML Aggarwal

 

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1 Answer

  1. Consider QR as the tower

    PQ as the pole on it

    Angle of elevation from P to a point A is ∠PAR = 600

    Angle of depression from Q to A = 450

    Here

    ∠QAR = 450 which is the alternate angle

    PQ = 5 m

    Take QR = h m

    PQ = 5 + h

    ML Aggarwal Solutions for Class 10 Chapter 20 Image 35

    In right triangle QAR

    tan θ = QR/AR

    Substituting the values

    tan 450 = h/AR

    So we get

    1 = h/AR

    AR = h

    In right triangle PAR

    tan 600 = PR/AR

    Substituting the values

    √3 = (5 + h)/ h

    So we get

    √3h = 5 + h

    h (√3 – 1) = 5

    h (1.732 – 1) = 5

    By further calculation

    0.732 h = 5

    h = 5/0.732 = 5000/732

    h = 6.83

    Hence, the height of the tower is 6.83 m.

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