QJEEM19Ma11a0: JEE Main 2019 Mathematics - Class 11 JEEM19Ma11a0 Solution: If the lengths of the sides of a triangle are in A.P. and the…

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