NCERT Exemplar Solutions
Class 10 - Mathematics - CHAPTER 8: Introduction to Trignometry and Its Applications - Exercise 8.4

Question 15

Step 1: Understand the setup

A ladder of fixed length \(L\) metres is leaning against a vertical wall. The bottom of the ladder (its foot) touches the ground, and the top touches the wall.

The angle between the ladder and the ground is initially \(\alpha\). After moving the foot of the ladder outward, the new angle with the ground is \(\beta\).

Step 2: Express the initial position

From right-angled triangle trigonometry:

• Base (horizontal distance from wall) = \(L \cos \alpha\) metres
• Height (vertical distance up the wall) = \(L \sin \alpha\) metres

Step 3: Express the final position

When the angle changes to \(\beta\):

• Base = \(L \cos \beta\) metres
• Height = \(L \sin \beta\) metres

Step 4: Calculate the change in base (\(p\))

The base has increased from \(L \cos \alpha\) to \(L \cos \beta\).

So, the change in base =

\[p = (L \cos \beta) - (L \cos \alpha) = L(\cos \beta - \cos \alpha)\]

Step 5: Calculate the change in height (\(q\))

The height has decreased from \(L \sin \alpha\) to \(L \sin \beta\).

So, the change in height =

\[q = (L \sin \alpha) - (L \sin \beta) = L(\sin \alpha - \sin \beta)\]

Step 6: Form the required ratio

We need to calculate \(\dfrac{p}{q}\).

Substitute the values of \(p\) and \(q\):

\[\dfrac{p}{q} = \dfrac{L(\cos \beta - \cos \alpha)}{L(\sin \alpha - \sin \beta)}\]

Step 7: Simplify

The ladder length \(L\) cancels out:

\[\dfrac{p}{q} = \dfrac{\cos \beta - \cos \alpha}{\sin \alpha - \sin \beta}\]

Final Result: The relation is proved.