Given, PQ = h metres, AB = a metres, ∠PAQ = 45°, ∠BAQ = 30°, ∠PBM = 60°
In rt. ΔAPQ, \(\frac{PQ}{AQ}\) = tan 45°
⇒ \(\frac{PQ}{AQ}\) = 1 ⇒ AQ = PQ = h
⇒ ΔAPQ is an isosceles rt. angled Δ
⇒ ∠APQ = ∠PAQ = 45°
∴ In rt. ΔPBM, ∠PMB = 90° and ∠PBM = 60°
⇒ ∠BPM = 30° ⇒ ∠APB = 45° – 30° = 15°.
Also ∠PAB = 15°.
∴ In ΔABP, ∠ABP = 180° – (∠APB + ∠PAB) = 180° – (15° + 15°) = 180° – 30° = 150°
Also, in ΔAPQ, AP = \(\sqrt{AQ^2+PQ^2}{}\) = \(\sqrt{h^2+h^2}\) = \(h\sqrt2\)
∴ By sine rule on ΔABP, \(\frac{AP}{\text{sin}\,150°}\) = \(\frac{AB}{\text{sin}\,15°}\) ⇒ \(\frac{h\sqrt2}{\frac{1}{2}}\) = \(\frac{a\sqrt2}{\frac{\sqrt3-1}{2\sqrt2}}\)
(Calculate sin 15° sin (60° – 45°) = sin 60° cos 45° – sin 45° cos 60°)
⇒ \(h\sqrt2\) = \(\frac{a\sqrt2}{\sqrt3-1}\) ⇒ a = h (\(\sqrt3-1\)).