Q. 45.0( 2 Votes )

# A man 1.8 metre t

Answer :

Let the height of tower be x metre and distance between tower and building be y metre.

Let CD be the height of man i.e. is 1.8 m. In right Δ ABD,
BD = (x + 1.8) m and AB = y m

BD = tan 60° × AB
(x + 1.8) = tan 60° × y
(From table, tan 60° = 1.732 )
(x + 1.8) = 1.732× y …(i)

In right Δ DEJ,
DJ = (1.8 + x – 10) m = (x – 8.2) m and EJ = AB = y m

DJ = tan 40° × EJ
(x – 8.2) = tan 40° × y
(From table, tan 40° = 0.839 = 0.84)
(x – 8.2) = 0.84× y …(ii)

Dividing eq. (i) from eq. (ii), 1.732(x – 8.2) = 0.84(x + 1.82)

1.732(x) – 1.732(8.2) = 0.84(x) + 0.84(1.82)

1.732(x) – 0.84(x) = 1.732(8.2) + 0.84(1.82)

x(1.732 – 0.84) = 14.2024 + 1.5288

x(0.892) = 15.7312 x = 17.63 m
Substituting the value of x in eq. (i)

(x + 1.8) = 1.732× y

(17.63 + 1.8) = 1.732(y)

19.43 = 1.732(y)

y = 11.21 m

Height of tower is 17.63 m and distance between tower and building is 11.21 m.

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