Question

Two buildings A and B are 100m apart The angle of elevation from the top of the building A to the top of the building B is 20° The angle of the elevation from the base of the building B to the top of the building A is 50° find the height of the building B

Answer 1

$Given, \: Distance \: between \: two$

$buildings \: A \: and \: B = 100 \: m$

$Height \: of \: the \: building \: A (AE) = x \:m$

$Height \: of \: the \: building \: B (BD) = (x + y)\:m$

$i ) In \: \triangle EAB , \angle { EAB} =90\degree$

$tan 50\degree = \frac{AE}{AB}$

$\implies 1.192 = \frac{x}{100}$

$\implies x = 1.192 \times 100$

$\implies x = 119.2 \: m\:---(1)$

$ii ) In \: \triangle ECD , \angle { ECD} =90\degree$

$tan 20\degree = \frac{CD}{CE}$

$\implies 0.3639 = \frac{y}{100}$

$\implies0.3639 \times 100 = y$

$\implies y = 0.3639 \times 100$

$\implies y =36.39 \: m\:---(2)$

$\therefore\red{ Height \: of \: the \: building \: B }$

$= BC + CD$

$= x + y$

$= 119.2 \: m + 36.39 \:m$

$= 155.59 \: m$

Therefore.,

$\therefore\red{ Height \: of \: the \: building \: B }$

$\green { = 155.59 \: m}$

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