Question

From a point 1.75 m above the ground and 10m away from a tower, the angle of elevation of the top of a tower is 60°. Find the height of the tower

Answer 1

Answer:

The answer to this question is....

Answer 2

$Given \: Height\:of \: a \: point \: from \: the$

$ground (AB) = 1.75 \:m$

$BC = 10 \:m$

$Let \: the \: height \: of \: the \: tower = CD$

$Angle \: of \: elevation = 60\degree$

$PD = x \: m$

$CP = BA = 1.75 \: m$

$In \: \triangle APD , \angle { APD} = 90\degree$

$tan \angle {DAP} = \frac{PD}{AP}$

$\implies tan 60\degree = \frac{x}{10}$

$\implies \sqrt{3} = \frac{x}{10}$

$\therefore x = 10\sqrt{3} \: m$

$x = 10 \times 1.732$

$\implies x = 17.32 \: m$

$\red{ Height \: of \: the \: Tower }$

$= CD$

$= CP + PD$

$= 1.75 + x$

$= 1.75 + 17.32$

$= 19.07 \: m$

Therefore.,

$\red{ Height \: of \: the \: Tower } \green { = 19.07 \: m }$

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