Answer
$8.50$
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, $A=180^{\circ}-70^{\circ}-60^{\circ}=50^{\circ}$
From formula $(c)$, we have:
$K=\dfrac{(4)^2\sin(70^{\circ})\sin(60^{\circ})}{(2) \sin(50^{\circ})} \approx 8.50$
