A man observes the angle of elevation of the top of a building to be 30°. He walks towards it in a horizontal line through its base. On covering 60 m, the angle of elevation changes to 60°. Find the height of the building correct to the nearest metre.

Let QR be the tower and man be initially at point P after moving 60 m let it reach point S.

PS = 60 m

From figure,

PQ = PS + SQ = (SQ + 60) m.     (Eq 1)

Considering right angled △SQR, we get

$\Rightarrow \text{tan 60°} = \dfrac{QR}{SQ} \\[1em] \Rightarrow \sqrt{3} = \dfrac{QR}{SQ} \\[1em] \Rightarrow QR = \sqrt{3} \text{ SQ} \text{ …..(Eq 2)}$

Considering right angled △PQR, we get

$\Rightarrow \text{tan 30°} = \dfrac{QR}{PQ} \\[1em] \Rightarrow \dfrac{1}{\sqrt{3}} = \dfrac{QR}{PQ} \\[1em] \Rightarrow PQ = \sqrt{3} \text{ QR} \\[1em] \Rightarrow SQ + 60 = \sqrt{3} \text{ QR} \text{ ……..(Eq 3)} \\[1em]$

Putting value of QR from Eq 2, in Eq 3 we get,

$\Rightarrow SQ + 60 = \sqrt{3} \times \sqrt{3} \text{ SQ} \\[1em] \Rightarrow SQ + 60 = 3 SQ \\[1em] \Rightarrow 2SQ = 60 \\[1em] \Rightarrow SQ = 30.$

From Eq 2,

$\Rightarrow QR = \sqrt{3}SQ \\[1em] \Rightarrow QR = \sqrt{3} \times 30 \\[1em] \Rightarrow QR = 51.96$

Correcting upto nearest meter QR = 52.

Hence, the height of building is 52 meters.