A ladder rests against a wall at angle \(\alpha\). Its foot is pulled away by \(p\) metres so that its top slides down \(q\) metres and now makes angle \(\beta\) with the ground. Prove that
\[\dfrac{p}{q}=\dfrac{\cos\beta-\cos\alpha}{\sin\alpha-\sin\beta}.\]
\(\dfrac{p}{q}=\dfrac{\cos\beta-\cos\alpha}{\sin\alpha-\sin\beta}\).
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:
Step 3: Express the final position
When the angle changes to \(\beta\):
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.