It would be easier to explain if the vertices were labeled.
Let's call the horizontal diagonal of the kite c.
The area of the triangle with sides [tex]5/6/c[/tex] triangle is given by
[tex]A_1 = \frac 1 2 ab\sin C = \frac 1 2 (5)(6)\sin 65 [/tex]
c itself is given by the Law of Cosines:
[tex]c^2 = 5^2 + 6^2 - 2(5)(6)\cos 65^\circ = 35.6429[/tex]
At this point I'd use a variation on Heron's formula to calculate the area. But the question says to use the sine and cosine rules so we'll do it that way.
The opposite angle, call it D, is given by the law of Cosines. We need its sine so we don't actually have to calculate the angle:
[tex]c^2 = 4^2 + 2.5^2 - 2(4)(2.5) \cos D[/tex]
[tex]\cos D = \dfrac{4^2 + 2.5^2 - 35.6429}{2(4)(2.5)} = -0.669645[/tex]
[tex]\sin D = \sqrt{1 - \cos^2 D} = 0.742681[/tex]
Now the other triangle has area
[tex]A_2 = \frac 1 2 (4)(2.5)(0.742681)[/tex]
So we get a total area of
[tex]A = A_1 + A_2 = \frac 1 2 (5)(6)\sin 65 + \frac 1 2 (4)(2.5)(0.742681) = 17.308[/tex]
