can you show the solution for this in finding the value of a? $$\frac{a^2+4^2-5^2}{8a} = \frac{\sqrt{3}}{2}\frac{a^2+4^2-6^2}{8a}+\frac{1}{2} \sqrt{1- \Big(\frac{a^2+4^2-6^2}{8a}\Big)^2}$$
show the solution in finding the  value of A

Let
$$\sin(\theta) = \frac{a^2 + 4^2 - 6^2}{8a}.$$
Then we have:
$$\sin(\theta) + \frac{11}{8a} = \frac{\sqrt{3}}{2}\sin(\theta) + \frac{1}{2}\cos(\theta).$$
Simplifying, we have:
$$\sin(\theta) + \frac{11}{8a} = \sin\left(\theta + \frac{\pi}{6}\right).$$
Consequently:
$$\sin\left(\theta + \frac{\pi}{6}\right) - \sin(\theta) = \frac{11}{8a}.$$
Simplifying the left-hand side of the last equation, we get:
$$\sqrt{2 - \sqrt{3}}\cos\left(\theta + \frac{\pi}{12}\right) = \frac{11}{8a}.$$
Therefore:
$$\cos\left(\theta + \frac{\pi}{12}\right) = \frac{11\left(\sqrt{2 + \sqrt{3}}\right)}{8a}.$$
Hence:
$$\theta = \arccos\left(\frac{11\left(\sqrt{2 + \sqrt{3}}\right)}{8a}\right) - \frac{\pi}{12}.$$
Now plug $\theta$ into:
$$\sin(\theta) = \frac{a^2 - 20}{8a},$$
and then solve for $a$.
Using WolframAlpha,
$$a \approx -0.798038 + 0.540401i,$$
or
$$a \approx -0.798038 - 0.540401i.$$
