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A person observes the angle of elevation of the top of a building as 30°. He proceeds towards the building with a speed of 25(√3 - 1)m/hour. After two hours, he observes the angle of elevation as 45°. What is the height of the building in metres?

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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.

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