Question

The angle of elevation and depression of the top and bottom of a tower from the top of a building 60m high are 30°and 60°

Answer 1

CORRECT QUESTION.

The angle of elevation and depression of the

top and the bottom of the tower from the top

of the building 60m high are 30° and 60°

respectively.Find the difference between the

heights of the building and the tower and the

distance between them.

EXPLANATION.

$\sf : \implies \: let \: we \: assume \: that \: ab \: be \: height \: of \: building. \\ \\ \sf : \implies \: let \: we \: assume \: that \: cd \: be \: the \: height \: of \: tower. \\ \\ \sf : \implies \: in \: right \: \triangle \: abd \\ \\ \sf : \implies \: \tan( \theta) = \frac{perpendicular}{base} \\ \\ \sf : \implies \: \tan(60 \degree) = \frac{ab}{bd}$

$\sf : \implies \: \sqrt{3} = \dfrac{60}{bd} \\ \\ \sf : \implies \: bd \: = \frac{60}{ \sqrt{3} } \\ \\ \sf : \implies \: bd \: = \frac{60}{ \sqrt{3} } \times \frac{ \sqrt{3} }{ \sqrt{3} } = \frac{60 \sqrt{3} }{3} \\ \\ \sf : \implies \: bd \: = 20 \sqrt{3}$

$\sf : \implies \: in \: right \triangle \: ace \\ \\ \sf : \implies \: \tan( \theta) = \frac{perpendicular}{base} \\ \\ \sf : \implies \: \tan(30 \degree) = \frac{ce}{ae} \\ \\ \sf : \implies \: \tan(30 \degree) = \frac{ce}{bd} \: \: (ae \: = bd) \\ \\ \sf : \implies \: \frac{1}{ \sqrt{3} } = \frac{ce}{20 \sqrt{3} } \\ \\ \sf : \implies \: ce \: = 20$

$\sf : \implies \: height \: of \: tower \: = ce \: + \: ed \: = 60 + 20 = 80 \: cm$

$\sf : \implies \: \orange{{1) = \underline{difference \: between \: the \: heigh \: of \: tower \: and \: building \: = 80 - 60 = 20 \: m}}} \\ \\ \sf : \implies \: \orange{{ 2) = \underline{distance \: between \: the \: tower \: and \: building \: = 20 \sqrt{3} }}}$