1

### 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

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

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

$\Rightarrow$  ${x^2} = 1200$  $\Rightarrow$ $x = 20\sqrt 3$
2

### 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.
3

### JEE Main 2019 (Online) 10th January Morning Slot

Consider a triangular plot ABC with sides AB = 7m, BC = 5m and CA = 6m. A vertical lamp-post at the mid point D of AC subtends an angle 30o at B. The height (in m) of the lamp-post is -
A
$2\sqrt {21}$
B
${3 \over 2}\sqrt {21}$
C
$7\sqrt {3}$
D
${2 \over 3}\sqrt {21}$

## Explanation BD = hcot30o = h$\sqrt 3$

So, 72 + 52 = 2(h$\sqrt 3$)2 + 32)

$\Rightarrow$  37 = 3h2 + 9

$\Rightarrow$  3h2 = 28

$\Rightarrow$  h = $\sqrt {{{28} \over 3}} = {2 \over 3}\sqrt {21}$
4

### JEE Main 2019 (Online) 10th January Evening Slot

With the usual notation, in $\Delta$ABC, if $\angle A + \angle B$ = 120o, a = $\sqrt 3$ $+$ 1, b = $\sqrt 3$ $-$ 1 then the ratio $\angle A:\angle B,$ is -
A
9 : 7
B
7 : 1
C
5 : 3
D
3 : 1

## Explanation

A + B = 120o $\tan {{A - B} \over 2} = {{a - b} \over {a + b}}\cot \left( {{c \over 2}} \right)$

$= {{\sqrt 3 + 1 - \sqrt 3 + 1} \over {2\left( {\sqrt 3 } \right)}}\cot \left( {{{30}^o}} \right) = {1 \over {\sqrt 3 }}.\sqrt 3 = 1$

${{A - B} \over 2} = {45^o}$

$\Rightarrow A - B = {90^o}$

$\ A + B = {120^o}$

$2A = {210^o}$

$A = {105^o}$

$B = {15^o}$

$\therefore$ $\angle A:\angle B,$ = 7 : 1