1

### 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.
2

### 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 = $\sqrt {{{28} \over 3}} = {2 \over 3}\sqrt {21}$
3

### 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
4

### JEE Main 2019 (Online) 11th January Morning Slot

In a triangle, the sum of lengths of two sides is x and the product of the lengths of the same two sides is y. If x2 – c2 = y, where c is the length of the third side of the triangle, then the circumradius of the triangle is :
A
${y \over {\sqrt 3 }}$
B
${c \over 3}$
C
${c \over {\sqrt 3 }}$
D
${3 \over 2}$y

## Explanation

Given a + b = x and ab = y

If x2 $-$ c2 = y $\Rightarrow$ (a + b)2 $-$ c2 = ab

$\Rightarrow$  a2 + b2 $-$ c2 = $-$ ab

$\Rightarrow$   ${{{a^2} + {b^2} - {c^2}} \over {2ab}} = - {1 \over 2}$

$\Rightarrow \cos C = - {1 \over 2}$

$\Rightarrow \angle C = {{2\pi } \over 3}$

$R = {c \over {2\sin C}} = {c \over {\sqrt 3 }}$