1. The problem involves finding the height of a tree using trigonometry. 2. Given: The distance from the instrument to the tree base is $9\sqrt{3}$ meters, and the angle of elevation to the top of the tree is $30^\circ$. 3. We use the formula for the height in a right triangle: $$\text{height} = \text{distance} \times \tan(\text{angle})$$ 4. Substitute the values: $$\text{height} = 9\sqrt{3} \times \tan(30^\circ)$$ 5. Recall that $\tan(30^\circ) = \frac{1}{\sqrt{3}}$. 6. Calculate: $$\text{height} = 9\sqrt{3} \times \frac{1}{\sqrt{3}} = 9$$ 7. Therefore, the height of the tree is $9$ meters. 1. Next, find the height of a car observed from the 4th floor with a depression angle of $60^\circ$ and horizontal distance $8\sqrt{3}$ meters. 2. Using the same formula for height (vertical distance): $$\text{height} = \text{distance} \times \tan(\text{angle})$$ 3. Substitute the values: $$\text{height} = 8\sqrt{3} \times \tan(60^\circ)$$ 4. Recall that $\tan(60^\circ) = \sqrt{3}$. 5. Calculate: $$\text{height} = 8\sqrt{3} \times \sqrt{3} = 8 \times 3 = 24$$ 6. Therefore, the height difference (vertical drop) is $24$ meters. Final answers: - Height of the tree: $9$ meters - Height difference to the car: $24$ meters