1. **Problem statement:** A man stands 50 m from the foot of a pole. His eye level is 1.5 m above the ground. The angle of elevation to the top of the pole is 30º. (1) Let the height of the pole be $x$ m. Find the length $AB$ in terms of $x$. (2) Find the height of the pole. (3) Find how much higher the pole is than the man's eye level. (4) If the man moves 33.34 m closer to the pole, find the new angle of elevation. Given $\tan 60^\circ = 1.732$. --- 2. **Step 1: Define variables and setup** - Let $AB$ be the vertical segment from the man's eye level to the top of the pole. - The total height of the pole is $x$ m. - The man's eye level is 1.5 m, so $AB = x - 1.5$. - The horizontal distance from the man to the pole is 50 m. 3. **Step 2: Use the tangent formula for angle of elevation** The tangent of the angle of elevation is the ratio of the opposite side to the adjacent side: $$\tan 30^\circ = \frac{AB}{50}$$ 4. **Step 3: Substitute $AB = x - 1.5$ and solve for $x$** $$\tan 30^\circ = \frac{x - 1.5}{50}$$ We know $\tan 30^\circ = \frac{1}{\sqrt{3}} \approx 0.577$. So, $$0.577 = \frac{x - 1.5}{50}$$ Multiply both sides by 50: $$x - 1.5 = 50 \times 0.577 = 28.85$$ Add 1.5 to both sides: $$x = 28.85 + 1.5 = 30.35$$ 5. **Step 4: Calculate how much higher the pole is than the man** Height difference: $$x - 1.5 = 30.35 - 1.5 = 28.85 \text{ m}$$ 6. **Step 5: New position and angle of elevation** The man moves 33.34 m forward, so new distance: $$50 - 33.34 = 16.66 \text{ m}$$ Using tangent again: $$\tan \theta = \frac{AB}{16.66} = \frac{28.85}{16.66} \approx 1.732$$ Given $\tan 60^\circ = 1.732$, so $$\theta = 60^\circ$$ --- **Final answers:** 1. $AB = x - 1.5$ 2. Height of pole $x = 30.35$ m 3. Pole is $28.85$ m higher than the man's eye level 4. New angle of elevation is $60^\circ$