Question #4395c

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Answer 1 · 2016-08-20T11:36:29

$\frac{h cos (α - β) sin α}{sin β}$ $(hcos(alpha-beta)sinalpha)/sinbeta$

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Given

$H \to Height of Tower = C D$ $H->"Height of Tower"=CD$
$h \to Height of Pole = A B$ $h->"Height of Pole"=AB$
$∠ C B D \to Angle of elevation top of Tower from B = α$ $/_CBD->"Angle of elevation top of Tower from B "=alpha$
$∠ A C B \to Angle subtended by the pole at C = β$ $/_ACB->"Angle subtended by the pole at C "=beta$
$∠ C A E = (α - β)$ $/_CAE =(alpha-beta)$
$Let A E = B D = b$ $"Let "AE=BD=b$

$"Now for "DeltaCBD ,(CD)/(BD)=H/b=tanalpha......(1)$

$"And for "DeltaCAE ,(CE)/(AE)=(H-h)/b=tan(alpha-beta)......(2)$

Dividing (2) by (1) we get

$(H-h)/H=tan(alpha-beta)/tanalpha$

$=>1-h/H=(sin(alpha-beta)cosalpha)/(cos(alpha-beta)sinalpha)$

$=>h/H=1-(sin(alpha-beta)cosalpha)/(cos(alpha-beta)sinalpha)$

$=>h/H=(cos(alpha-beta)sinalpha-sin(alpha-beta)cosalpha)/(cos(alpha-beta)sinalpha)$

$=>h/H=sin(alpha-alpha+beta)/(cos(alpha-beta)sinalpha)$

$=>h/H=sinbeta/(cos(alpha-beta)sinalpha)$

$=>H/h=(cos(alpha-beta)sinalpha)/sinbeta$

$=>H=(hcos(alpha-beta)sinalpha)/sinbeta$