Question

A bridge across a valley is h meters long. There is a temple in the valley directly below the bridge. The angles of depression of the top of the temple from the two ends of the bridge are
$\alpha \: and \: \beta$Prove that the height of the bridge above the top of the temple is
$h{tan \ \: \alpha tan \: \beta } \div tan \: \alpha + tan \: \beta$
meters

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Answer 1

$\large\underline\purple{\bold{Solution :- }}$

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☆ Let the distance between bridge and top of the temple be 'y' meters.

⇛ DC = y,

☆ where C is point on AB such that DC is perpendicular to AB.

☆ Let AB be the length of the bridge, which is 'h' meters.

☆ Let D be the top of the temple.

☆ According to the statement,

☆ ∠BAD=α and ∠ABD=β

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$\tt \:  ⟼ In \: right \: \triangle \: ACD$

$\tt \:  ⟼ tan \alpha = \dfrac{DC}{AC}$

$\tt : \implies \:tan \alpha = \dfrac{y}{x}$

$\tt : \implies \:x = \dfrac{y}{tan \alpha } - - - - (1)$

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$\tt \:  ⟼ In \: right \: \triangle \: BCD$

$\tt \:  ⟼ tan \beta = \dfrac{DC}{BC}$

$\tt : \implies \:tan \beta = \dfrac{y}{h - x}$

$\tt : \implies \:h - x = \dfrac{y}{tan \beta }$

☆ On substituting the value of 'x' from equation (1), we get

$\tt : \implies \:h - \dfrac{y}{tan \alpha } = \dfrac{y}{tan \beta }$

$\tt : \implies \:h = \dfrac{y}{tan \alpha } + \dfrac{y}{tan \beta }$

$\tt : \implies \:h = \dfrac{ytan \beta + ytan \alpha }{tan \alpha \: tan \beta }$

$\tt : \implies \:h = \dfrac{y(tan \beta + tan \alpha) }{tan \alpha \: tan \beta }$

$\tt : \implies \:y = \dfrac{h \: tan \alpha \: tan \beta }{tan \beta \: + \: tan \alpha }$

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$\large{\boxed{\boxed{\bf{Hence, Proved}}}}$

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