Answer
$1.89$
Work Step by Step
We know that the area of a triangle with sides $a,b$ and $c$ and their opposite angles $A, B$ and $C$ can be represented as:
$K=\dfrac{a^2sin(B)\sin(C)}{2\sin(A)}; \\ K=\dfrac{b^2\sin(A)\sin(C)}{2\sin(B)}; \\ K=\dfrac{c^2\sin(A)\sin(B)}{2\sin(C)}$
Remember that the sum of the angles of the triangle is $180^{\circ}$
Therefore, $B=180^{\circ}-50^{\circ}-20^{\circ}=110^{\circ}$
Now, we have: $K=\dfrac{a^2sin(B)\sin(C)}{2\sin(A)}=\dfrac{(3)^2\sin(110^{\circ})\sin(20^{\circ})}{2\sin(50^{\circ})} \approx 1.89$