Area Of A Triangle!

Let $BC = x$ ; then $AB = 2x$ . Let $\angle BAC = \theta$ . By sine rule ,

$\begin{cases} \dfrac {AB}{BD} = \dfrac {\sin \angle ADB}{\sin \angle BAD} & \implies \dfrac {2x}a = \dfrac {\sin (120^\circ - \theta)}{\sin \theta} & ...(1) \\ \dfrac {BC}{BD} = \dfrac {\sin \angle BDC}{\sin \angle BCD} & \implies \dfrac xa = \dfrac {\sin (60^\circ + \theta)}{\sin (60^\circ -\theta)} & ...(2) \end{cases}$

$\begin{aligned} \frac {(1)}{(2)}: \quad 2 & = \frac {\blue{\sin(120^\circ - \theta)}}{\sin \theta} \times \frac {\sin(60^\circ - \theta)}{\sin (60^\circ+\theta)} & \small \blue{\text{Note that }\sin (180^\circ - \phi) = \sin \phi} \\ & = \frac {\blue{\sin(60^\circ + \theta)}}{\sin \theta} \times \frac {\sin(60^\circ - \theta)}{\sin (60^\circ+\theta)} \\ & = \frac {\sin(60^\circ - \theta)}{\sin \theta} \\ & = \frac {\sqrt 3}2 \cot \theta - \frac 12 \end{aligned}$

$\implies \tan \theta = \dfrac {\sqrt 3}5 \implies \sin \theta = \dfrac {\sqrt{21}}{14}, \cos \theta = \dfrac {5\sqrt 7}{14}$ and from $(1)$ :

$\begin{aligned} \frac {2x}a & = \frac {\sin (60^\circ + \theta)}{\sin \theta} = \frac {\frac {\sqrt 3}2 \times \frac {5\sqrt 7}{14} + \frac 12 \times \frac {\sqrt{21}}{14}}{\frac {\sqrt {21}}{14}} = 3 \\ \implies x & = \frac 32 a \\ A_{\triangle ABC} & = \frac 12 AB \times BD \sin 60^\circ + \frac 12 BC \times BD \sin 60^\circ \\ & = \frac 12 (2x+x) a \left(\frac {\sqrt 3}2\right) \\ & = \frac {9\sqrt 3}8 a^2 = 11250 \sqrt 3 \\ \implies a & = \sqrt{\frac {8 \times 11250}9} = \boxed{100} \end{aligned}$

Using the Law of sines for $\triangle{ABD}$ we have:

$\dfrac{2}{\sqrt{3}}AD = \dfrac{2y}{\sin(180^{\circ} - m)} = \dfrac{2y}{\sin(m)} \implies$ $AD = \dfrac{\sqrt{3}y}{\sin(m)}$

and using the Law of sines for $\triangle{DBC}$ we have:

$\dfrac{2}{\sqrt{3}}DC = \dfrac{y}{\sin(m)} \implies DC = \dfrac{\sqrt{3}y}{2\sin(m)}$

$\implies \dfrac{AD}{DC} = 2$ .

Let $DC = x \implies AD = 2x$ .

Using the Law of cosines for $\triangle{ABD}$ we have:

$4x^2 = 4y^2 + a^2 - 4ay\cos(60^{\circ}) = 4y^2 - 2ay + a^2$

and using the Law of cosines for $\triangle{DBC}$ we have:

$x^2 = y^2 + a^2 - 2ay\cos(60^{\circ}) = y^2 - ay + a^2$

Solving the two equations:

$4x^2 = 4y^2 - 2ay + a^2$

$-4 (x^2 = y^2 - ay + a^2)$

$\implies 2ay - 3a^2 = 0 \implies a(2y - 3a) = 0$ and $a \neq 0 \implies y = \dfrac{3a}{2}$

$\implies x^2 = \dfrac{9}{4}a^2 - \dfrac{3}{2}a^2 + a^2 = \dfrac{7}{4}a^2 \implies x = \dfrac{\sqrt{7}}{2}a$

$\implies DC = \dfrac{\sqrt{7}}{2}a = \dfrac{\sqrt{3}(\dfrac{3}{2})a}{2\sin(m)} \implies$ $\sin(m) = \dfrac{3}{2}\sqrt{\dfrac{3}{7}}$ and $BD = a \implies$

the height $h$ of $\triangle{ABC}$ is $h = \dfrac{3}{2}\sqrt{\dfrac{3}{7}}a$

and $AC = a \implies A_{\triangle{ABC}} = \dfrac{1}{2}(\dfrac{3\sqrt{7}}{2})(\dfrac{3}{2})\sqrt{\dfrac{3}{7}} a^2 = \dfrac{9}{8}\sqrt{3} a^2 = 11250\sqrt{3} =$

$2 * 25^2 * 3^2\sqrt{3} \implies a^2 = 2^4 * (25)^2 \implies a = \boxed{100}$ .