The angle of elevation of the top of a tree from a point 27m away and on the same horizontal ground as the foot of the tree is 30\(^o\). Find the height of ...

We can solve this problem using trigonometry, specifically the tangent function. Let's draw a diagram to visualize the situation. We have a right triangle with the height of the tree as one of the legs, the distance from the tree to the point on the ground as the other leg, and the angle of elevation (30 degrees) as the angle opposite the height of the tree.  We can use the tangent function to find the height of the tree: $$\tan(30^\circ) = \frac{\text{height of tree}}{27\text{ m}}$$ We know that the tangent of 30 degrees is equal to 1/\(\sqrt{3}\) (or approximately 0.577), so we can substitute that in and solve for the height of the tree: $$\frac{1}{\sqrt{3}} = \frac{\text{height of tree}}{27\text{ m}}$$ Multiplying both sides by 27 m gives: $$\text{height of tree} = \frac{27\text{ m}}{\sqrt{3}} = 9\sqrt{3}\text{ m}$$ Therefore, the height of the tree is 9\(\sqrt{3}\) meters. Answer option (D) is correct.