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.
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(90∘−α)=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