Answer:
in right angle triangle ABD relationship between base and hypotenuse is given by cos angle
Cos A=[tex] \frac{base}{hypotenuse} [/tex]
A=[tex]{cos }^{ - 1} (\frac{12}{8 \sqrt{3} } ) = {30}^{0} [/tex]
again
in right angle triangle ACD
relationship between base and hypotenuse is given by cos angle
Cos 30°=[tex] \frac{base}{hypotenuse} = \frac{8 \sqrt{3} }{x} [/tex]
[tex] \frac{ \sqrt{3} }{2} = \frac{8 \sqrt{3} }{x} \\ x = \frac{8 \times 2 \sqrt{3} }{ \sqrt{3} } = 16[/tex]
x=16 is your answer.
