Q. 35.0( 5 Votes )

# A 1.5 metre tall

Let x m be actual height of building and BD be the height of

boy i.e. BD = 1.5 m.

Let the distance between the boy and the building be y m.

In right Δ DFC,
FC = AC – AF = (x – 1.5) m and FD = AB = y m

FC = tan 60° × FD
(x – 1.5) = tan 60° × y
(From table, tan 60° = 1.732 )
(x – 1.75) = 1.732× y …(i)

In right Δ DFG,
FG = AC – CG – AF = (x – 10 – 1.5) m = (x – 11.5) m
and FD = AB = y m

FG = tan 30° × FD

(x – 11.5) = tan 30° × y
(From table, tan 30° = 0.57)

(x – 11.5) = 0.57 × y …(ii)

Dividing eq. (i) from eq. (ii)

1.732(x – 11.5) = 0.57(x – 1.75)

1.732(x) – 1.732(11.5) = 0.57(x) – 0.57(1.75)

1.732(x) – 0.57(x) = 1.732(11.5) – 0.57(1.75)

(1.732 – 0.57)x = 19.918 – 0.9975

1.162(x) = 18.92

Height of the building is 16.28 m.

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