# Finding the area of the given quadrilateral

In the given quadrilateral, we know the angles are as written on diagram $60,60,90,150$ degrees and the given sides have length: $6 , 4\sqrt3$. I want to find the area by transforming it into two triangles. Note that I know a method is extending the leftmost side and making a greater equilateral triangle and then subtracting extra parts. I don't want this method. I want to calculate the area by decomposing the shape into two triangles. I want to know how I can find the angles of these triangles in order to find the area. Suppose $\angle B$ is subdvided by the dotted line into $\angle B_1$ and $\angle B_1$ and suppose the dotted line was $q$. Now, we have this system of equations: $$\frac{\sin B_1}{6}=\frac{\sin 60}{q}, \sin B_2=\frac{4\sqrt{3}}q\Rightarrow q=q$$ $$\Rightarrow\frac{6\sin 60}{\sin B_1}=\frac{4\sqrt3}{\sin B_2}; B_1+B_2=60^\circ$$ Now substituting $B_1=60-B_2$, we get: $$\frac{6\sin60}{\sin(60-B_2)}=\frac{4\sqrt3}{\sin B_2}$$ Which can be further simplified into: $$\frac{6\cdot\frac{\sqrt3}{2}}{\frac{\sqrt3}{2}\cos B_2-\frac12\sin B_2}=\frac{4\sqrt3}{\sin B_2}$$ $$\Rightarrow6\sqrt3\sin B_2+4\sqrt3\sin B_2=12\cos B_2$$ Dividing both sides by $\cos B_2$ we get: $$10\sqrt3\tan B_2=12$$ $$B_2=\tan^{-1}\frac{12}{10\sqrt3}\approx34.71^\circ$$ $$B_1=60^\circ-\tan^{-1}\frac{12}{10\sqrt3}\approx25.29^\circ$$ Now recall that $q=\frac{6\sin60}{\sin B_1}$, thus: $$q=\frac{6\sin60}{\sin\Big(60-\tan^{-1}\frac{12}{10\sqrt3}\Big)}=\sqrt{148}$$
Now knowing that $\triangle ABD$ is a right triangle, it can be easily shown that $AB=10$.Thus the area of $ABD$ is: $$A_{\triangle ABD}=\frac12 q\cdot AB\sin B_2=20\sqrt3$$
Now to find the area of the $\triangle BCD$, we need to find $\angle D$ in that triangle, thus: $$\angle D=180^\circ-60^\circ-B_1\approx=94.71^\circ$$ And thus, the area of $\triangle BCD$ is: $$A_{\triangle BCD}=\frac12 q\cdot DC\sin D=21\sqrt3$$ So the area of the quadrilateral is: $$\therefore\bbox[15px, border: 2px solid black]{A_{ABCD}=A_{\triangle ABD}+A_{\triangle BCD}=20\sqrt3+21\sqrt3=41\sqrt3}$$
Using sine rule on $\triangle DAB$ and $\triangle CDB$ $$\frac{DB}{\sin(90^\circ)} = \frac{AD}{\sin(\angle ABD^\circ)}= \frac{4\sqrt3}{\sin(\angle ABD^\circ)}\\ \frac{DB}{\sin(60^\circ)} = \frac{CD}{\sin(\angle DBC^\circ)}= \frac{6}{\sin(60^\circ-\angle ABD^\circ)}$$
So we get $\angle ABD$ and $DB$ from here. Rest is straight forward.