Question

the ladder is 8 M long reaches a distance 8m from the top of the vertical flagstaff .At the foot of the ladder the elevation of the top of the flagstaff is 60° find the height of the flagstaff

Answer 1

Answer:

$\red { Length \: of \: flagship }\green {= 12\:m}$

Step-by-step explanation:

$Given \: Length \:of \: the \: ladder (AC) = 8m$

$CD = 8 \:m$

$\angle BAD = 60\degree$

$Let \: BC = h \: m$

$AB = x \:m$

$In \: \triangle ABD ,\\\angle BAD + \angle B + \angle D = 180\degree$

$\blue {( Angle \: sum \: Property )}$

$\implies \angle 60\degree + 90\degree + \angle D = 180\degree$

$\implies \angle D = 180\degree - 150\degree \\= 30\degree$

$\angle BAD = 60\degree$

$\implies \angle BAC + \angle CAD = 60\degree$

$\implies \angle BAC + \angle 30\degree = 60\degree$

$\implies \angle BAC = 60 - 30 = 30\degree$

$In \: \triangle BAD , \\tan \angle BAD = \frac{BD}{AB}$

$\implies tan \:60\degree = \frac{BD}{BA}$

$\implies \sqrt{3} = \frac{8+h}{x}$

$\implies x = \frac{8+h}{\sqrt{3}}\: --(1)$

$In \: \triangle ABC , \\tan \:\angle BAC = \frac{BC}{BA}$

$\implies tan \:30\degree = \frac{h}{x}$

$\implies \frac{1}{\sqrt{3}} = \frac{h}{x}$

$\implies x = \sqrt{3} h \: ---(2)$

$[ From \: (1) \:and \:(2) ]$

$\implies \frac{8+h}{\sqrt{3}} = \sqrt{3} h$

$\implies 8+h = 3h$

$\implies 8 = 3h - h$

$\implies 8 = 2h$

$\implies \frac{8}{2} = h$

$\implies 4 = h$

Therefore.,

$\red { Length \: of \: flagship }$

$= h + 8 \\= 4 + 8 \\= \green { 12\:m }$

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