Question

On a
4
feet pillar an antenna of
8
feet is erected. The angle of depression from the top of the pillar to a

point
A
on the ground is
53°
and the angle of depression from the top of the antenna to a point
B on the ground is 45°. Find AB.

Answer 1

$Height \:of \:the \:Pillar (QR) = 4 \:ft$

$Height \:of \:the \:Attenna (PQ) = 8 \:ft$

$Distance \:from \:foot \:of \:the \:Pillar \\to \:point \:A = x \:ft$

$Distance \:from \:foot \:of \:the \:Pillar \\to \:point \:B= (x + y )\:ft$

$i ) In \: \triangle PBR , \:we \:have$

$tan 45 \degree = \frac{PR}{BR}$

$\implies 1 = \frac{12}{x+y}$

$\implies x + y = 12 \: ---(1)$

$ii ) In \: \triangle QAR , \:we \:have$

$tan 53 \degree = \frac{QR}{AR}$

$\implies 1.33 = \frac{4}{x}$

$\implies x = \frac{4}{1.33}$

$\implies x = 3.01 \:ft\: ---(2)$

/* Substitute x = 3.01 in equation (1) ,we get */

$\implies 3.01 + y = 12$

$\implies y = 12 - 3.01 = 8.99\:ft$

Therefore.,

$\red { Distance \: of \:AB \:(y) } \green {= 8.99 \:ft }$

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