1

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

### 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
3

### 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

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

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

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

$R = {c \over {2\sin C}} = {c \over {\sqrt 3 }}$
4

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

Given ${{b + c} \over {11}} = {{c + a} \over {12}} = {{a + b} \over {13}}$ for a $\Delta$ABC with usual notation.

If   ${{\cos A} \over \alpha } = {{\cos B} \over \beta } = {{\cos C} \over \gamma },$ then the ordered triad ($\alpha$, $\beta$, $\gamma$) has a value
A
(19, 7, 25)
B
(7, 19, 25)
C
(5, 12, 13)
D
(3, 4, 5)

## Explanation

b + c = 11$\lambda$, c + a = 12$\lambda$, a + b = 13$\lambda$

$\Rightarrow$  a = 7$\lambda$, b = 6$\lambda$, c = 5$\lambda$

(using cosine formula)

cosA = ${1 \over 5},$ cosB = ${19 \over 35},$ cosC = ${5 \over 7},$

$\alpha$ : $\beta$ : $\gamma$ $\Rightarrow$  7 : 19 : 25