NCERT Exemplar Solutions
Class 10 - Mathematics - CHAPTER 8: Introduction to Trignometry and Its Applications - Exercise 8.2

Question 12

Let the original height of the tower be \(h\) metres, and the distance of the point of observation from the foot of the tower be \(d\) metres.

The angle of elevation \(\theta\) is given by the formula:

\[ \tan \theta = \dfrac{\text{opposite side}}{\text{adjacent side}} = \dfrac{h}{d} \]

Now, if both the height and the distance are increased by 10%:

• New height = \(h + 10\%\,\text{of}\,h = 1.1h\) metres
• New distance = \(d + 10\%\,\text{of}\,d = 1.1d\) metres

The new angle of elevation is \(\theta'\), and:

\[ \tan \theta' = \dfrac{1.1h}{1.1d} \]

Simplify the fraction:

\[ \tan \theta' = \dfrac{1.1}{1.1} \times \dfrac{h}{d} = \dfrac{h}{d} \]

This means:

\[ \tan \theta' = \tan \theta \]

So, \(\theta' = \theta\).

Therefore, the angle of elevation remains unchanged even if both the height and the distance are increased by the same percentage.