An observer 1.6 m tall is $20\sqrt{3}$ m away from a tower. The angle of elevation from his eye to the top of the tower is $30^\circ$. The height of the tower is :

### Concept & Two-Part Height Calculation When an observer has a stated height, the trigonometric calculation only gives the height of the object *above the observer's eye level*. The observer's height must be added to find the total height. $$ \text{Total Height} = (\text{Distance} \times \tan(\theta)) + \text{Observer Height} $$ ### Step-by-Step Solution 1. Given: Observer height = $1.6$ m, Distance to tower ($d$) = $20\sqrt{3}$ m, Angle of elevation ($\theta$) = $30^\circ$. 2. Let the height of the tower above the observer's eye level be $x$. 3. Using the right triangle formed above eye level: $$ \tan(30^\circ) = \frac{x}{20\sqrt{3}} $$ 4. Substitute $\tan(30^\circ) = \frac{1}{\sqrt{3}}$: $$ \frac{1}{\sqrt{3}} = \frac{x}{20\sqrt{3}} $$ 5. Solve for $x$: $$ x = \frac{20\sqrt{3}}{\sqrt{3}} = 20 \text{ m} $$ 6. Add the observer's height to get the total tower height: $$ \text{Total Height} = 20 \text{ m} + 1.6 \text{ m} = 21.6 \text{ m} $$ ### Exam Strategy & Shortcut Visualize the $30^\circ$-$60^\circ$-$90^\circ$ triangle. If the base ($20\sqrt{3}$) corresponds to the $\sqrt{3}$ side, the opposite side (height above eye) must be $20$. Add $1.6$ to get $21.6$ instantly without writing equations. ### Common Pitfall Forgetting to add the observer's height ($1.6$ m) to the calculated height ($20$ m) and falsely answering $20$ m if it were an option. ### Final Answer Therefore, the correct answer is **21.6 m**.