Question

From the top of a building 60m high, the angle of elevation of the top of a vertical pole is 15°. at the bottom of the building the angle of elevation of the top of the pole is 35°. find (a) the height of the pole and (b) the distance of the pole from the building

Answer 1

Height of the building (AB) = 60m,

Height of the pole (CE) = CD + DE

= ( 60 + x ) m ,

Distance of the pole from the building (BC)= d m,

BC = AD = d m,

<A = 15° , <EBC = 35°

$i ) In \: \triangle ADE ,\\tan \: 15\degree = \frac{ED}{AD}$

$\implies (2-\sqrt{3}) = \frac{x}{d}$

$\implies (2-1.732) = \frac{x}{d}$

$\implies 0.268 = \frac{x}{d}$

$\implies x = 0.268d \: ---(1)$

$ii ) In \: \triangle BCD ,\\tan \: 35\degree = \frac{EC}{BC}$

$\implies 0.7 = \frac{x + 60}{d}$

$\implies 0.7d = x + 60$

$\implies 0.7d = 0.268d + 60 \: [ From \: (1) ]$

$\implies 0.7d - 0.268d = 60$

$\implies 0.432d = 60$

$\implies d = \frac{60}{0.432}$

$\implies d = 138.9 \: m\: ---(2)$

/* Put value of d in equation (1) , we get */

$x = 0.268 \times 138.9$

$\implies x = 37.2 \:m$

$Height \: of \: the \: Pole (CE) = x + 60 \\= 37.2\:m + 60\:m \\= 97.2 \:m$

Therefore.,

$\red { Height \: of \: the \: Pole} \green { =97.2 \:m }$

$\red {Distance of the pole from the building }\green { = 37.2 \:m }$

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