Let's break this problem down and solve it step by step:

Part a) Drawing a rough figure

We can visualize the problem by drawing two right-angled triangles representing the two different observations by the boy.

1. Initial position: The boy is standing at a point where he sees the top of the tower at an elevation of $60^\circ$. Let’s call the horizontal distance between this position and the base of the tower as $d_1$.

2. Stepping back 20 meters: The boy steps back 20 meters, and the elevation to the top of the tower decreases to $30^\circ$. Let the new distance from the tower be $d_2 = d_1 + 20$.

The height of the tower is the same in both triangles, and the boy's height (1 meter) is factored in.

Part b) Finding the height of the tower

Let’s assume:

• $h_T$ is the height of the tower.
• $d_1$ is the initial horizontal distance from the tower.

We have two right-angled triangles:

• From the first observation at $60^\circ$, we can use the trigonometric relation: $\tan(60^\circ) = \frac{h_T - 1}{d_1}$ This gives us: $\sqrt{3} = \frac{h_T - 1}{d_1}$ $h_T - 1 = d_1\sqrt{3} \quad \text{(Equation 1)}$

• From the second observation at $30^\circ$, the relation is: $\tan(30^\circ) = \frac{h_T - 1}{d_1 + 20}$ $\frac{1}{\sqrt{3}} = \frac{h_T - 1}{d_1 + 20}$ $h_T - 1 = \frac{d_1 + 20}{\sqrt{3}} \quad \text{(Equation 2)}$

Solving for $d_1$ and $h_T$:

From Equation 1 and Equation 2, we have two expressions for $h_T - 1$. Equating them gives: $d_1\sqrt{3} = \frac{d_1 + 20}{\sqrt{3}}$ Multiplying both sides by $\sqrt{3}$ to eliminate the fraction: $3d_1 = d_1 + 20$ $2d_1 = 20$ $d_1 = 10 \, \text{meters}$

Now, substitute $d_1 = 10$ back into Equation 1 to find the height of the tower: $h_T - 1 = 10\sqrt{3}$ $h_T = 1 + 10\sqrt{3} = 1 + 17.32 = 18.32 \, \text{meters}$

Thus, the height of the tower is approximately 18.32 meters.

Do you need further clarifications or details?

Here are five related questions to deepen your understanding:

1. How would the calculations change if the boy's height were different?
2. What would happen to the angle of elevation if the boy moved even further back?
3. How can we verify the answer using trigonometric identities?
4. How do different trigonometric functions relate to angle and distance in such problems?
5. Can this problem be solved graphically as well?

Tip: Always double-check your trigonometric calculations, especially when working with different angles and distances.