<s>[INST] What is the area of the triangle here? 
We are given that the angle of $BAD$ is $2\alpha$ and the angle of $DAC$ is $\alpha$. $|AC| = 10$, $|BD| = 6$, $|DC| = 5$ units. Find the area of the triangle $ABC$.
The answer would be $33$. We need to show that if we drop an altitude from $A$ to $BC$ at point $E$, $|AE| = 6, |EC| = 8$. Somehow $\triangle AEC$ becomes $6$-$8$-$10$ triangle. Since $|AE| = 6$ and $|BC|=11$, the area becomes $33$.
But how can we prove that?
 [/INST] Let area of $\triangle BAD = A_1$ and area of $\triangle CAD = A_2$. Let $AD = m$ and $AB = x$.
As they share a common height, the ratio $\frac{A_1}{A_2} = \frac 65$
Also, using an alternative formula for triangular area, $\frac{A_1}{A_2} = \frac{\frac 12xm\sin 2\alpha}{\frac 12(m)(10)\sin \alpha} = \frac{x\cos\alpha}{5}$ (where a double angle identity was used).
Equating the two expressions and rearranging, $\cos \alpha = \frac 6x$.
Now apply cosine rule on $\triangle ABC$, to give $11^2 = x^2 + 10^2 - 2(x)(10)\cos 3\alpha$.
Using a triple angle identity ($\cos 3\alpha = 4\cos^3 \alpha - 3\cos\alpha$),
$x^2 - 20x(4(\frac 6x)^3 - 3(\frac 6x)) - 21 = 0$
After a bit of algebra and setting $y = x^2$ to reduce the quartic to a quadratic, we get:
$y^2 + 339y - 17280 = 0$.
Solving for the positive root, $y = 45 \implies x  =\sqrt{45}$.
At this point, we can proceed to find the area of $\triangle ABC$ in a couple of ways. One way is to apply Heron's formula using the three known sides. I instead opted to apply the previously used formula, so that the area $A = \frac 12(x)(10)\sin 3\alpha$.
From $\cos \alpha = \frac 6x$ we get $\sin \alpha = \frac 3{\sqrt{45}}$. Applying a triple angle formula ($\sin 3\alpha = 3\sin\alpha - 4\sin^3\alpha$),
$A = \frac 12(x)(10)(\frac 9{\sqrt{45}} - 4(\frac{27}{45\sqrt{45}})) = 5\sqrt{45}(\frac 9{\sqrt{45}} - \frac{12}{5\sqrt{45}}) = 45- 12 = 33$.
So the answer is $33$.
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