In the adjoining figure, from the top of a building AB, 60 metres high

1.	In the adjoining figure, from the top of a building AB, 60 metres high, the angles of depression of the top and the bottom of a vertical lamppost CD are observed to be 30 degree and 60 degree respectively. Find1. the horizontal distance between AB and CD.2. the height of the lamppost CD.
Answer» ANSWER: $\setlength{\unitlength}{1cm}\begin{picture}(0,0)\thicklines\put(0,0){\line(0,1){6}}\put(0,0){\line(1,0){4}}\put(4,0){\line(0,1){4}}\multiput(0,4)(0.6,0){7}{\qbezier(0,0)(0,0)(0.5,0)}\qbezier(0,6)(0,6)(4,0)\qbezier(0,6)(0,6)(4,4)\put(0,6){\vector(1,0){3}}\qbezier(0.8,6)(1,5.8)(0.6,5.7)\put(1,5.6){$30^{\circ}$}\qbezier(1.8,6)(2.3,5)(1,4.6)\put(2.2,5.2){$60^{\circ}$}\qbezier(3.2,4.42)(3,4.1)(3.25,4)\put(2.4,4.15){$30^{\circ}$}\qbezier(3.3,0)(3.3,0.5)(3.7,0.5)\put(-0.5,-0.5){B}\put(2.7,0.3){$60^{\circ}$}\put(-0.5,6){A}\put(-0.5,3.8){E}\put(-0.5,5){x}\put(4.3,-0.5){D}\put(4.3,3.8){C}\end{picture}$ $\displaystyle\underline{\bigstar\:\textsf{Horizontal Distance :}}$ $\displaystyle\sf :\implies tan \ \theta = \dfrac{Opposite}{Adjacent}\\\\$ $\displaystyle\sf :\implies tan \ 60^{\circ} = \dfrac{<klux>AB</klux>}{BD}\\\\$ $\displaystyle\sf :\implies \sqrt{3} = \dfrac{AB}{BD}\\\\$ $\displaystyle\sf :\implies BD = \dfrac{AB}{\sqrt{3}}\\\\$ $\displaystyle\sf :\implies BD = \dfrac{60}{\sqrt{3}}\\\\$ $\displaystyle:\implies\textsf{ \textbf{ BD = 20}}\sf{ \sqrt{3}}\textsf{\textbf{m}}$ $\displaystyle\underline{\bigstar\:\textsf{<klux>HEIGHT</klux> of the <klux>LAMPPOST</klux> :}}$ $\displaystyle\sf \dashrightarrow tan \ \theta = \dfrac{Opposite}{Adjacent}\\\\$ $\displaystyle\sf \dashrightarrow tan \ 30^{\circ} = \dfrac{<klux>AE</klux>}{EC}\\\\$ $\displaystyle\sf \dashrightarrow \dfrac{1}{\sqrt{3}} = \dfrac{x}{20\sqrt{3}}\\\\$ $\displaystyle\sf \dashrightarrow x = 20$ AB - AE will be the height of the lamppost $\displaystyle\sf \dashrightarrow 60-x\\$ $\displaystyle\sf \dashrightarrow 60-20\\$ $\displaystyle\sf \dashrightarrow\textsf{ \textbf{ BE = 40 m}}$

In the adjoining figure, from the top of a building AB, 60 metres high, the angles of depression of the top and the bottom of a vertical lamppost CD are observed to be 30 degree and 60 degree respectively. Find1. the horizontal distance between AB and CD.2. the height of the lamppost CD.

Answer»

ANSWER:

$\setlength{\unitlength}{1cm}\begin{picture}(0,0)\thicklines\put(0,0){\line(0,1){6}}\put(0,0){\line(1,0){4}}\put(4,0){\line(0,1){4}}\multiput(0,4)(0.6,0){7}{\qbezier(0,0)(0,0)(0.5,0)}\qbezier(0,6)(0,6)(4,0)\qbezier(0,6)(0,6)(4,4)\put(0,6){\vector(1,0){3}}\qbezier(0.8,6)(1,5.8)(0.6,5.7)\put(1,5.6){$30^{\circ}$}\qbezier(1.8,6)(2.3,5)(1,4.6)\put(2.2,5.2){$60^{\circ}$}\qbezier(3.2,4.42)(3,4.1)(3.25,4)\put(2.4,4.15){$30^{\circ}$}\qbezier(3.3,0)(3.3,0.5)(3.7,0.5)\put(-0.5,-0.5){B}\put(2.7,0.3){$60^{\circ}$}\put(-0.5,6){A}\put(-0.5,3.8){E}\put(-0.5,5){x}\put(4.3,-0.5){D}\put(4.3,3.8){C}\end{picture}$

$\displaystyle\underline{\bigstar\:\textsf{Horizontal Distance :}}$

$\displaystyle\sf :\implies tan \ \theta = \dfrac{Opposite}{Adjacent}\\\\$

$\displaystyle\sf :\implies tan \ 60^{\circ} = \dfrac{<klux>AB</klux>}{BD}\\\\$

$\displaystyle\sf :\implies \sqrt{3} = \dfrac{AB}{BD}\\\\$

$\displaystyle\sf :\implies BD = \dfrac{AB}{\sqrt{3}}\\\\$

$\displaystyle\sf :\implies BD = \dfrac{60}{\sqrt{3}}\\\\$

$\displaystyle:\implies\textsf{ \textbf{ BD = 20}}\sf{ \sqrt{3}}\textsf{\textbf{m}}$

$\displaystyle\underline{\bigstar\:\textsf{<klux>HEIGHT</klux> of the <klux>LAMPPOST</klux> :}}$

$\displaystyle\sf \dashrightarrow tan \ \theta = \dfrac{Opposite}{Adjacent}\\\\$

$\displaystyle\sf \dashrightarrow tan \ 30^{\circ} = \dfrac{<klux>AE</klux>}{EC}\\\\$

$\displaystyle\sf \dashrightarrow \dfrac{1}{\sqrt{3}} = \dfrac{x}{20\sqrt{3}}\\\\$

$\displaystyle\sf \dashrightarrow x = 20$

AB - AE will be the height of the lamppost

$\displaystyle\sf \dashrightarrow 60-x\\$

$\displaystyle\sf \dashrightarrow 60-20\\$

$\displaystyle\sf \dashrightarrow\textsf{ \textbf{ BE = 40 m}}$