Area of a Quadrilateral!

For $\triangle{ADC} :$

$\dfrac{\sin(5x)}{DC} = \dfrac{\sin(8x)}{AC} \implies \dfrac{\sin(5x)}{\sin(8x)} = \dfrac{DC}{AC} = \dfrac{BC}{AC} = \dfrac{1}{AC}$

For $\triangle{ABC} :$

$\dfrac{\sin(3x)}{BC} = \dfrac{\sin(4x)}{AC} \implies \dfrac{\sin(3x)}{\sin(4x)} = \dfrac{BC}{AC} = = \dfrac{1}{AC}$

$\implies$

$\dfrac{\sin(5x)}{\sin(8x)} = \dfrac{\sin(3x)}{\sin(4x)} \implies \sin(5x)\sin(4x) = \sin(8x)\sin(3x) =$

$2\sin(4x)\cos(4x)\sin(3x) \implies \sin(5x) = 2\cos(4x)\sin(3x)$

Using the identity $2\cos(A)\sin(B) = \sin(A + B) - \sin(A - B)$ , with $A = 4x$ and

$B = 3x$ we obtain: $\sin(5x) = \sin(7x) - \sin(x) \implies \sin(7x) - \sin(5x) = \sin(x)$

or $\sin(7x) + \sin(-5x) = \sin(x)$

Using the identity $\sin(A) + \sin(B) = 2\sin(\dfrac{A + B}{2})\cos(\dfrac{A - B}{2}),$ with $A = 7x$ and

$B = -5x$ we obtain: $2\sin(x)\cos(6x) = \sin(x) \implies 2\cos(6x) = 1 \implies \cos(6x) = \dfrac{1}{2}$

$\implies 6x = 60^{\circ} \implies \boxed{x = 10^{\circ}}$

Using the above diagram with $x = 10^{\circ} \implies h_{2} = \sin(30^{\circ}) = \dfrac{1}{2}$ and $h_{1} = \sin(10^{\circ})$

For $AC$ we have: $\dfrac{AC}{\sin(140^{\circ})} = \dfrac{1}{\sin(30^{\circ})} = 2$ $\implies AC = 2\sin(140^{\circ})$

$\implies$ The total area $A_{ABCD} = \dfrac{1}{2}(2\sin(140^{\circ})\sin(10^{\circ})) + \dfrac{1}{2}(2\sin(140^{\circ}))(\dfrac{1}{2}) =$

$\sin(140^{\circ})(\sin(10^{\circ}) + \dfrac{1}{2}) = \dfrac{1}{2}\sin(140^{\circ})(2\sin(10^{\circ}) + 1)$ $\approx \boxed{0.43301270}$

From the angle conditions (solutions above showed that $x=10^{\circ}$ ), we know that $AB||DC$ and so the quadrilateral is a trapezium, the height of the trapezium is $1\times \sin (180^{\circ}-\angle ABC)=\sin 40^{\circ}$ .

By sine rule, $AB=\sin 10^{\circ}\times \frac{1}{\sin 30^{\circ}}=2\sin 10^{\circ}.$ So the area of quadrilateral $ABCD$ is $\frac{1}{2}(1+2\sin 10^{\circ})(\sin 40^{\circ})\approx \boxed{0.43301270}$

By sine rule :

$\begin{aligned} \frac {AC}{BC} & = \frac {AC}{CD} = \frac {AC}1 \\ \frac \blue{\sin (180^\circ - 4x)}{\sin 3x} & = \frac \blue{\sin (180^\circ - 8x)}{\sin 5x} & \small \blue{\text{SInce } \sin(180^\circ - \theta) = \sin \theta} \\ \frac \blue{\sin 4x}{\sin 3x} & = \frac \blue{\sin 8x}{\sin 5x} & \small \blue{\text{And } \sin 2\theta = 2 \sin \theta \cos \theta} \\ \frac \blue 1{\sin 3x} & = \frac \blue{2\cos 4x}{\sin 5x} \\ \frac {\sin 5x}{\sin 3x} & = 2 \blue{\cos 4x} & \small \blue{\text{Also }\sin \theta = \cos (90^\circ - \theta)} \\ \frac \blue{\sin 5x}\red{\sin 3x} & = \frac \blue{\sin (90^\circ - 4x)}\red{\frac 12} \\ \implies \blue{5x} & = \blue{90^\circ - 4x} \implies x = 10^\circ \\ \implies \red{3x} & = \red{30^\circ} \implies x = 10^\circ \end{aligned}$

Fill in the measures of all the angles. Extend $BC$ and $DA$ to meet at $E$ . Then we note that $\angle E = 40^\circ$ and $\triangle CDE$ and $\triangle ABE$ are isosceles and similar. Therefore the area of quadrilateral $ABCD$ :

$\begin{aligned} [ABCD] & = [CDE]-[ABE] \\ & = \frac 12 \cdot 1^2 \cdot \sin 80^\circ - \frac 12 \cdot AB^2 \cdot \sin 80^\circ & \small \blue{\text{By sine rule, }AB = \frac {\sin 10^\circ}{\sin 30^\circ}} \\ & = \frac {\sin 80^\circ}2 \left(1 - \blue{\frac {\sin^2 10^\circ}{\sin^2 30^\circ}} \right) \\ & \approx \boxed{0.433} \end{aligned}$