I WANT TO SOLVE THIS TRIANGLE

A little tricky idea can solve it amaizingly. Since the area of the triangle is 2 sqaure units, so let's assume the $\angle A = 90 ^ \circ$ and $b=c=2$ [ so that the area is $\frac{1}{2} \times 2 \times 2 = 2$ ].

Now, since we assumed $b=c=2$

Therefore, $\angle B = \angle C= 45^ \circ$

$\Rightarrow c^{2}sin2B + b^{2}sin2C= 2b^{2}\sin (2B) =2 \times 2^{2} \times1= 8$

area of triangle=2(square units);

c^2 sin2B + b^2 Sin2C=?

tools-

sin(x+y)=sinxcosy+cosxsiny

a/SinA = b/SinB=c/SinC

=) SinC=c SinB/b

=) SinB=bSinC/c

Sin2B=2sinBcosB

sin2C=2SinCcosC

now the solving part

c^2 Sin2B +b^2Sin2C

=c^2 (2sinBcosB) + b^2 (2SinCcosC)

=2(c^2(bSinCcosB/c) + b^2 (c SinBcosC/b))

=2cb(sinCcosB+sinBcosC)

=2cbSin(C+B)--------(Sin (C+B)=Sin (180-(c+b))=Sin(A)

=2cbSinA----eq 1

Area of the triangle = bcSinA/2 =2

=) bcSinA=4

so value of eq 1= 4*2 = 8