Question

A straight highway leads to the foot of a tower. A man standing at the top of

the tower observes a car at an angle of depression of 30°, which is approaching

the foot of the tower with a uniform speed. Six seconds later, the angle of

depression of the car is found to be 60°. Find the time taken by the car to

reach the foot of the tower from this point.​

Answer 1

$\underline{\frak{Diagram:}}$

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$\setlength{\unitlength}{1.9cm}\begin{picture}(6,2)\linethickness{0.4mm}\put(8,1){\line(1,0){4}}\put(8,1){\line(0,2){1.9}}\qbezier(10.5,1)(10,1.4)(8,2.9)\qbezier(12,1)(11,1.4)(8,2.9)\put(7.5,2){\sf{\large{h}}}\put(8.2,1){\line(0,1){0.2}}\put(8,1.2){\line(3,0){0.2}}\qbezier(9.8,1)(9.7,1.25)(10,1.4)\qbezier(11,1)(10.8,1.2)(11.1,1.4)\put(9.6,1.2){\sf\large{60^\circ }}\put(10.2,1.3){\sf\footnotesize{30^ \circ }}\put(11.9,.7){\sf\large B}\put(7.9,3){\sf\large A}\put(10.4,.7){\sf\large C}\put(7.9,.7){\sf\large D}\put(8.1,0.8){\vector(3,0){0.8}}\put(10.3,0.8){\vector( - 3,0){0.8}}\put(8.95,0.75){\sf ? m}\put(8.1,0.4){\vector(3,0){1.6}}\put(12,0.4){\vector( - 3,0){1.6}}\put(9.8,0.35){\sf }\end{picture}$⠀

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Given: Man standing at the top of the tower observes a car at an angle of depression of 30°. Six seconds later, angle of depression of the card is found to be 60°. Height of the tower is AD.

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Need to find: We've to find out CD.

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$\bigstar\:{\underline{\sf{In\: right\:angled\:\triangle\:ACD,}}}$

$:\implies\sf tan \; C = \dfrac{AD}{CD} \\\\\\:\implies\sf tan\;60^{\circ} = \dfrac{AD}{CD} \\\\\\:\implies\sf 30^{\circ} = \dfrac{AD}{CD}\\\\\\:\implies\sf \sqrt{3} CD = AD\\\\\\:\implies\sf AD = \sqrt{3} CD \qquad\quad\bigg\lgroup\bf Equation\;(I)\bigg\rgroup$

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$\bigstar\:{\underline{\sf{In\: right\:angled\:\triangle\:ABD,}}}$

$:\implies\sf tan\;60^{\circ} = \dfrac{AD}{BD}\\\\\\:\implies\sf \dfrac{1}{\sqrt{3}} = \dfrac{AD}{BD}\\\\\\:\implies\sf \dfrac{BD}{\sqrt{3}} = AD\\\\\\:\implies\sf AD = \dfrac{BD}{\sqrt{3}}\qquad\quad\bigg\lgroup\bf Equation \;(II)\bigg\rgroup$

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$\underline{\bf{\dag} \:\mathfrak{Comparing\;both\; equations\: :}}$⠀⠀⠀⠀⠀⠀⠀⠀

$:\implies\sf \sqrt{3}CD = \dfrac{BD}{\sqrt{3}}\\\\\\:\implies\sf \sqrt{3}\Big(\sqrt{3} CD\Big) = BD\\\\\\:\implies\sf 3CD = BD$

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• With the help of diagram we can see that BD is (BC + CD). Therefore,

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$:\implies\sf 3CD = BC + CD\\\\\\:\implies\sf 3CD - CD = BC \\\\\\:\implies\sf 2CD = BC\\\\\\:\implies\sf CD = \dfrac{BC}{2}\qquad\quad\bigg\lgroup\bf Equation\;(III)\bigg\rgroup$

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$\underline{\bf{\dag} \:\mathfrak{By \;using\; equation\; (3)\: :}}$⠀⠀⠀⠀

• Time to cover BC is given which is 6 seconds. Now,

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$:\implies\sf CD = \dfrac{BC}{2}\\\\\\:\implies\sf CD = \cancel\dfrac{6}{2}\\\\\\:\implies{\underline{\boxed{\frak{\pink{CD = 3}}}}}\;\bigstar$

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$\therefore{\underline{\sf{Hence, \; required\;time\;taken\;is\;\bf{ 3\; seconds}.}}}$

Answer 2

Answer :-

Let height be AO

And triangle be ABC

So

tan C = AO/CO

Tan 60 = AO/CO

$\sf \sqrt{3} CO = AO$

$\sf AO = \sqrt{3}CO$

Now

tan 60 = A0/BO

$\frac{1}{\sqrt{3}} = \frac{AO}{BO}$

AO = CO/$\sf \sqrt{3}$

on comparing

$\sf \sqrt{3} CO = \frac{BD}{\sqrt{3} }$

$\sf 3CO = BO$

Now

3CO = BC + CO

3CO - CD = BC

2CO = BC

CO = BC/2

Now

CD = 6/2 = 3 SEC