1

### JEE Main 2016 (Online) 10th April Morning Slot

The angle of elevation of the top of a vertical tower from a point A, due east of it is 45o. The angle of elevation of the top of the same tower from a point B, due south of A is 30o. If the distance between A and B is $54\sqrt 2 \,m,$ then the height of the tower (in metres), is :
A
$36\sqrt 3$
B
54
C
$54\sqrt 3$
D
108

## Explanation

Let the height of tower = h

In triangle PQA,

tan45o = ${h \over {QA}}$

$\Rightarrow$   h = QA

In triangle PQB,

tan300 = ${h \over {BQ}}$

$\Rightarrow$   BQ = $\sqrt 3$h

In triangle BAQ

$\angle$ QAB = 90o.

$\therefore$    QA2 + AB2 = QB2

$\Rightarrow$   h2 + (54$\sqrt 2$)2 = ($\sqrt 3$h)2

$\Rightarrow$   2h2 = (54$\sqrt 2$)2

$\Rightarrow$   $\sqrt 2 h$ = 54$\sqrt 2$

$\Rightarrow$   h = 54 m
2

### JEE Main 2018 (Online) 15th April Morning Slot

An aeroplane flying at a constant speed, parallel to the horizontal ground, $\sqrt 3$ kmabove it, is obsered at an elevation of ${60^o}$ from a point on the ground. If, after five seconds, its elevation from the same point, is ${30^o}$, then the speed (in km / hr) of the aeroplane, is :
A
1500
B
1440
C
750
D
720

## Explanation

For $\Delta$OA, A, OA1 = ${{\sqrt 3 } \over {\tan {{60}^o}}}$ = 1 km

For $\Delta$OB1, B, OB1 = ${{\sqrt 3 } \over {\tan {{30}^o}}}$ = 3km

As, a distance of 3 $-$ 1 = 2 km is convered in 5 seconds.

Therefore the speed of the plane is

${{2 \times 3600} \over 5}$ = 1440 km/hr
3

### JEE Main 2018 (Online) 15th April Evening Slot

A tower T1 of height 60 m is located exactly opposite to a tower T2 of height 80 m on a straight road. Fromthe top of T1, if the angle of depression of the foot of T2 is twice the angle of elevation of the top of T2, then the width (in m) of the road between the feetof the towers T1 and T2 is :
A
$10\sqrt 2$
B
$10\sqrt 3$
C
$20\sqrt 3$
D
$20\sqrt 2$

## Explanation

Let the distance between T1 and T2 be x

From the figure

EA = 60 m (T1) and DB = 80 m (T2)

$\angle DEC = \theta$ and $\angle BEC = 2\theta$

Now in $\angle DEC$,

$\tan \theta = {{DC} \over {AB}} = {{20} \over x}$

and in $\Delta BEC$,

$\tan 2\theta = {{BC} \over {CE}} = {{60} \over x}$

We know that

$\tan 2\theta = {{2\tan \theta } \over {1 - {{\left( {\tan \theta } \right)}^2}}}$

$\Rightarrow$  ${{60} \over x} = {{2\left( {{{20} \over x}} \right)} \over {1 - {{\left( {{{20} \over x}} \right)}^2}}}$

$\Rightarrow$  ${x^2} = 1200$  $\Rightarrow$ $x = 20\sqrt 3$
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### JEE Main 2018 (Online) 16th April Morning Slot

A man on the top of a vertical tower observes a car moving at a uniform speed towards the tower on a horizontal road. If it takes 18 min. for the angle of depression of the car to change from 30o to 45o ; then after this, the time taken (in min.) by the car to reach the foot of the tower, is :
A
$9\left( {1 + \sqrt 3 } \right)$
B
$18\left( {1 + \sqrt 3 } \right)$
C
$18\left( {\sqrt 3 - 1} \right)$
D
${9 \over 2}\left( {\sqrt 3 - 1} \right)$

## Explanation

Assume height of tower = AB = h

From $\Delta$ABD,

tan45o = ${{AB} \over {AD}}$

$\Rightarrow $$\,\,\, {{AB} \over {AD}} = 1 [ as tan45o = 1] \Rightarrow \,\,\, AB = AD \therefore\,\,\, AD = h From \Delta BAC, tan30o = {{AB} \over {AC}} \Rightarrow \,\,\, {h \over {AC}} = {1 \over {\sqrt 3 }} \Rightarrow \,\,\, AC = h \sqrt 3 As, AC = AD + DC \Rightarrow$$\,\,\,$ DC = AC $-$ AD = $\sqrt 3 h$ $-$ h

Given that, time taken to reach from point C to D = 18 min.

$\therefore\,\,\,$ Car speed = ${{distance} \over {time}}$ = ${{CD} \over {18}}$ = ${{(\sqrt 3 - 1)h} \over {18}}$

$\therefore\,\,\,$ Time taken to move from D to A

= ${{Distan ce\,\,of\,\,DA} \over {speed}}$

= ${h \over {{{\left( {\sqrt 3 - 1} \right)h} \over {18}}}}$

= ${{18} \over {\left( {\sqrt 3 - 1} \right)}}$

= ${{18\left( {\sqrt 3 + 1} \right)} \over {3 - 1}}$

= 9 $\left( {\sqrt 3 + 1} \right)$ min.