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Rajan@2021
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The angles of elevation of the top of a tower from two points at a distance of 4 m and 9 m from the base of the tower and in the same straight line with it are complementary. Prove that the height of the tower is 6 m

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This is the basic and conceptual question from height and distance in which we are to Show that the height of the tower is 6 m,  if the angles of elevation of the top of a tower from two points at a distance of 4 m and 9 m from the base of the tower and in the same straight line with it are complementary.

RS Aggarwal, Class 10, chapter 14, question no 29.

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  1. Given AB is the tower.
    P and Q are the points at distance 4m and 9m respectively.
    From the figure, PB=4m and QB=9m
    Let angle of elevation from P be α and angle of elevation from Q be β
    Given that α and β are supplementary.
    α+β=90
    In ABP,tanα=AB​/BP            …..(1)
    In ABQ,tanβ=AB/BQ
    tan(90α)=AB/BQ since α+β=90
    cotα=AB​/BQ
    ⇒1/tanα=AB/BQ
    So, tanα=BQ​/AB                         …….(2)
    From (1) and (2) we have
    AB​/BP=BQ/AB
    AB^2=BP.BQ=4×9=36
    AB=6m
    Hence, height of the tower=6m

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