Let the height of the building PQ be h metres.
Let R be the first point of observation such that ∠PRQ = 30°.
Let S be the second point of observation such that ∠PSQ = 45°
RS = Distance moved in two hours with a speed of 25\((\sqrt3-1)\)m/jpir
= 2 x 25\((\sqrt3-1)\)m = 50\((\sqrt3-1)\) m
Let SQ = \(x\) metres
Then, in ΔPQS, tan 45° = \(\frac{PQ}{QS}\) ⇒ \(\frac{h}{x}\) = 1 ⇒ h = \(x\) .....(i)
In ΔPQR, we have tan 30° = \(\frac{PQ}{RQ}\) ⇒ \(\frac{h}{x50(\sqrt3-1)}\) = \(\frac{1}{\sqrt3}\)
⇒ √3 h = \(x\) + 50 \((\sqrt3-1)\) ⇒ √3 h = h + 50 \((\sqrt3-1)\) (From (i) \(x\) = h)
\((\sqrt3-1)\)h = 50 \((\sqrt3-1)\) ⇒ h = 50 m.
∴ Height of the building = 50 metres.