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