Give the justification of the construction also:
Draw a triangle $ABC$ with side $BC = 7 \, cm$,$\angle B = 45^{\circ}$,$\angle A = 105^{\circ}$. Then,construct a triangle whose sides are $\frac{4}{3}$ times the corresponding sides of $\triangle ABC$.

(A) $\angle B = 45^{\circ}, \angle A = 105^{\circ}$
The sum of all interior angles in a triangle is $180^{\circ}$.
$\angle A + \angle B + \angle C = 180^{\circ}$
$105^{\circ} + 45^{\circ} + \angle C = 180^{\circ}$
$\angle C = 180^{\circ} - 150^{\circ} = 30^{\circ}$
The required triangle can be drawn as follows:
$1.$ Draw a $\triangle ABC$ with side $BC = 7 \, cm$,$\angle B = 45^{\circ}$,$\angle C = 30^{\circ}$.
$2.$ Draw a ray $BX$ making an acute angle with $BC$ on the opposite side of vertex $A$.
$3.$ Locate $4$ points $B_1, B_2, B_3, B_4$ on $BX$ such that $BB_1 = B_1B_2 = B_2B_3 = B_3B_4$.
$4.$ Join $B_3C$. Draw a line through $B_4$ parallel to $B_3C$ intersecting the extended $BC$ at $C'$.
$5.$ Through $C'$,draw a line parallel to $AC$ intersecting the extended line segment $BA$ at $A'$. $\triangle A'BC'$ is the required triangle.
Justification:
In $\triangle ABC$ and $\triangle A'BC'$,
$\angle ABC = \angle A'BC'$ (Common)
$\angle ACB = \angle A'C'B$ (Corresponding angles)
$\triangle ABC \sim \triangle A'BC'$ ($AA$ similarity criterion)
$\Rightarrow \frac{AB}{A'B} = \frac{BC}{BC'} = \frac{AC}{A'C'} \dots(1)$
In $\triangle BB_3C$ and $\triangle BB_4C'$,
$\angle B_3BC = \angle B_4BC'$ (Common)
$\angle BB_3C = \angle BB_4C'$ (Corresponding angles)
$\triangle BB_3C \sim \triangle BB_4C'$ ($AA$ similarity criterion)
$\Rightarrow \frac{BC}{BC'} = \frac{BB_3}{BB_4} = \frac{3}{4} \dots(2)$
From $(1)$ and $(2)$,we get $\frac{AB}{A'B} = \frac{BC}{BC'} = \frac{AC}{A'C'} = \frac{3}{4}$.
Thus,$A'B = \frac{4}{3} AB$,$BC' = \frac{4}{3} BC$,and $A'C' = \frac{4}{3} AC$. This justifies the construction.