1

### 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

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

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

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

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

### 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
3

### JEE Main 2019 (Online) 12th January Evening Slot

If the angle of elevation of a cloud from a point P which is 25 m above a lake be 30o and the angle of depression of reflection of the cloud in the lake from P be 60o, then the height of the cloud (in meters) from the surface of the lake is :
A
45
B
42
C
50
D
60

## Explanation tan 30o = ${x \over y} \Rightarrow y = \sqrt 3 x\,\,\,\,....(i)$

tan 60o = ${{25 + x + 25} \over y}$

$\Rightarrow \,\,\sqrt 3 y = 50 + x$

$\Rightarrow$  $3x = 50 + x$

$\Rightarrow$   x = 25 m

$\therefore$   Height of cloud from surface

= 25 + 25 = 50m
4

### JEE Main 2019 (Online) 8th April Evening Slot

If the lengths of the sides of a triangle are in A.P. and the greatest angle is double the smallest, then a ratio of lengths of the sides of this triangle is :
A
5 : 9 : 13
B
5 : 6 : 7
C
4 : 5 : 6
D
3 : 4 : 5

## Explanation

Let smallest angle $\angle A$ = $\theta$

and largest angle $\angle C$ is double of $\angle A$.

$\therefore$ $\angle C$ = 2$\theta$

$\therefore$ $\angle B$ = $\pi$ - 3$\theta$

Let length of sides are

a, b, c

where a < b < c

As a, b, c are in A.P then

2b = a + c

From sin rule we can say,

2 sin B = sin A + sin C

$\Rightarrow$ 2 sin ($\pi$ – 3$\theta$) = sin $\theta$ + sin 2$\theta$

$\Rightarrow$ 2 sin (3$\theta$) = sin $\theta$ + sin 2$\theta$

$\Rightarrow$ 2(3sin $\theta$ - 4sin3 $\theta$) = sin$\theta$ + 2sin $\theta$ cos $\theta$

$\Rightarrow$ 2(3 - 4sin2 $\theta$) = 1 + 2cos $\theta$

$\Rightarrow$ 2(3 - 4 + 4cos2 $\theta$) = 1 + 2cos $\theta$

$\Rightarrow$ 2(4cos2 $\theta$ - 1) = 1 + 2cos $\theta$

$\Rightarrow$ 8 cos2 $\theta$ – 2 cos $\theta$ – 3 = 0

$\Rightarrow$ 8 cos2 $\theta$ – 6 cos $\theta$ + 4 cos $\theta$ – 3 = 0

$\Rightarrow$ (2 cos $\theta$ + 1) ( 4 cos $\theta$ - 3) = 0

$\Rightarrow$ cos $\theta$ = ${3 \over 4}$ or cos $\theta$ = $-{1 \over 2}$

But cos $\theta$ = $-{1 \over 2}$ is not possible as $\angle C$ is acute angle.

Now we have to find,

a : b : c

$\Rightarrow$ sin A : sin B : sin C [ from sin rule]

$\Rightarrow$ sin $\theta$ : sin ($\pi$ - 3$\theta$) : sin 2$\theta$

$\Rightarrow$ sin $\theta$ : sin 3$\theta$ : sin 2$\theta$

$\Rightarrow$ sin $\theta$ : (3sin $\theta$ - 4sin3 $\theta$) : 2sin $\theta$ cos $\theta$

$\Rightarrow$ 1 : (3 - 4sin2 $\theta$) : 2 cos $\theta$

$\Rightarrow$ 1 : (3 - 4 + 4cos2 $\theta$) : 2 cos $\theta$

$\Rightarrow$ 1 : (4cos2 $\theta$ - 1) : 2 cos $\theta$

$\Rightarrow$ 1 : (4${\left( {{3 \over 4}} \right)^2}$ - 1) : 2 $\times$ ${{3 \over 4}}$

$\Rightarrow$ 1 : ${5 \over 4}$ : ${{3 \over 2}}$

$\Rightarrow$ 4 : 5 : 6