Answer
$2.27$
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:
$a) K=\dfrac{a^2sin(B)\sin(C)}{2\sin(A)}; \\ b) K=\dfrac{b^2\sin(A)\sin(C)}{2\sin(B)}; \\ c) K=\dfrac{c^2\sin(A)\sin(B)}{2\sin(C)}$
Remember that the sum of the angles of the triangle is $180^{\circ}$
Therefore, $A=180^{\circ}-70^{\circ}-10^{\circ}=100^{\circ}$
From formula $(b)$, we have: $K=\dfrac{(5)^2\sin(100^{\circ})\sin(10^{\circ})}{(2) \sin(70^{\circ})} \approx 2.27$