Easy trigo

Consider the diagram above.

$\tan \theta=\dfrac{h}{8}=\dfrac{DC}{h}$ $\implies$ $\dfrac{h}{8}=\dfrac{DC}{h}$ $\implies$ $h^2=8DC$

Apply pythagorean theorem on triangle $ADC$ , $h^2=(4\sqrt{3})^2-(DC)^2=48-(DC)^2$

$h^2=h^2$

$48-(DC)^2=8DC$

$(DC)^2+8DC-48=0$

$(DC+12)(DC-4)=0$

$DC=-12$ or $DC=4$ (disregard the negative value)

Therefore,

$h^2=8DC=8(4)=32$

$h=\sqrt{32}=\sqrt{16(2)}=$ $\boxed{4\sqrt{2}}$

Let $DC = x$

Now,notice,

$\triangle ADC \sim \triangle BAC$

Therefore,

$\frac{ AC }{ BC } = \frac{ DC }{ AC }$

$\Rightarrow \frac {4 \sqrt{3} }{ 8 + x } = \frac{ x }{ 4\sqrt{3} }$

$\Rightarrow 48 = x^2 + 8x$

Solving the equation.we get,

$x =$ -12 , 4

Since length of side cannot be negative, So,

DC = 4

Now, in $\triangle ADC$

$AD^2 = AC^2 - DC^2$

$\Rightarrow AD^2 = 48 - 16 = 32$

$\Rightarrow AD = \sqrt{32} = \boxed{4 \sqrt{2} }$

From Leg Corollary to the Right Angle Altitude Theorem, (8+DC)/AB=AB/BD. AB= sqrt(8+y)^2-(4sqrt3)^2 from Phythagorean Theorem. Hence, DC= 4 From Hypotenuse Corollary to the Right Traingle Altitide Theorem, BD/AD=AD/DC. Therefore, 8/AD=AD/4 AD= 4sqrt2