Question

Someone please help me ASAP!!! I'll give Brainliest and TONS OF POINTS!!!

Answer 1

Answer:

in right angle triangle ABD relationship between base and hypotenuse is given by cos angle

Cos A=$\frac{base}{hypotenuse}$

A=${cos }^{ - 1} (\frac{12}{8 \sqrt{3} } ) = {30}^{0}$

again

in right angle triangle ACD

relationship between base and hypotenuse is given by cos angle

Cos 30°=$\frac{base}{hypotenuse} = \frac{8 \sqrt{3} }{x}$

$\frac{ \sqrt{3} }{2} = \frac{8 \sqrt{3} }{x} \\ x = \frac{8 \times 2 \sqrt{3} }{ \sqrt{3} } = 16$

x=16 is your answer.