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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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