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Let D C = x
Now,notice,
△ A D C ∼ △ B A C
Therefore,
B C A C = A C D C
⇒ 8 + x 4 3 = 4 3 x
⇒ 4 8 = x 2 + 8 x
Solving the equation.we get,
x = -12 , 4
Since length of side cannot be negative, So,
DC = 4
Now, in △ A D C
A D 2 = A C 2 − D C 2
⇒ A D 2 = 4 8 − 1 6 = 3 2
⇒ A D = 3 2 = 4 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
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tan θ = 8 h = h D C ⟹ 8 h = h D C ⟹ h 2 = 8 D C
Apply pythagorean theorem on triangle A D C , h 2 = ( 4 3 ) 2 − ( D C ) 2 = 4 8 − ( D C ) 2
h 2 = h 2
4 8 − ( D C ) 2 = 8 D C
( D C ) 2 + 8 D C − 4 8 = 0
( D C + 1 2 ) ( D C − 4 ) = 0
D C = − 1 2 or D C = 4 (disregard the negative value)
Therefore,
h 2 = 8 D C = 8 ( 4 ) = 3 2
h = 3 2 = 1 6 ( 2 ) = 4 2