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(C) Let the height of the tower be $H$ and the height of the building be $h$. The horizontal distance $d = 10 \sqrt{3}$.From the top of the tower,the

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$40$

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(C) Let the height of the tower be $H$ and the height of the building be $h$. The horizontal distance $d = 10 \sqrt{3}$.From the top of the tower,the angle of depression to the foot of the building is $60^{\circ}$. Thus,$	an(60^{\circ}) = \frac{H}{d}$.$H = d 	imes 	an(60^{\circ}) = 10 \sqrt{3} 	imes \sqrt{3} = 10 	imes 3 = 30$ units.From the foot of the tower,the angle of elevation to the top of the building is $30^{\circ}$. Thus,$	an(30^{\circ}) = \frac{h}{d}$.$h = d 	imes 	an(30^{\circ}) = 10 \sqrt{3} 	imes \frac{1}{\sqrt{3}} = 10$ units.The sum of the heights is $H + h = 30 + 10 = 40$ units.

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