In a narrow, sinister alley of width *w* a ladder of length *a* is placed with its foot at point P between the walls. Resting against one wall at Q, a distance *k* above the ground, the ladder makes a 45° angle with the ground. Resting against the other wall at R, a distance *h* above the ground, the ladder makes a 75° angle with the ground. The width of the alley is equal to

*a*- RQ
*k*- (
*h*+*k*)/2 *h*

- Draw QE ⊥ QB and AR.
- Draw QR; △PQR is isosceles.
- ∠RPQ = 180° – 75° – 45° = 60° → △PQR is equilateral → ∠PRQ = 60°.
- ∠ARP = 90° – 75° = 15° → ∠EQR = 90° – (60° + 15°) = 15°.
- △ERQ ≅ △APR, by ASA (∠ERQ = 75°).
*h*= AR = EQ = AB =*w*. The answer is (e).