Pointy Quadrilateral

The hypotenuse $BD=\sqrt{3^2+4^2}=5$

Also $13^2=12^2+5^2\implies\angle BDC=90^\circ$

$[ABCD]=[ABD]+[BCD]=\frac12\times3\times4+\frac12\times12\times5=6+30=\boxed{36}$

Connecting B and D creates a right triangle with BD the hypotenuse. The length of BD is then 5, being a 3-4-5 right triangle. The angle at C can be found using the law of cosines: $5^{2}=13^{2}+12^{2}-2(12)13cos(C)$ , therefore $C = arccos(\frac{25-144-169}{-312})$ $= arccos(\frac{12}{13})$ $= 22.61986495^{o}$ . The semiperimeter: $s = \frac{4+3+13+12}{2} = 16$ . Using Bretschneider's Formula the area of the quadrilateral can be found to be: $\sqrt{(16-13)(16-12)(16-4)(16-3)-13(12)4(3)cos^{2}(.5(90^{o}+22.61986495^{o})})$ = $\sqrt{1872-1872cos^{2}(56.30993247)} = \sqrt{1296} = 36$

I was unable to find suitable methods of simplifying the cos(.5(90+arccos(12/13)) by hand, and as such had to resort to a calculator. If anyone can offer insght - maybe using a power series approximation?