Two poles of heights 7m and 8m stand on the ground. The distance between their feet is x meters. From the top of the 7m pole, the angle of elevation to the top of the 8m pole is 30 degrees. Find the d

Let the height of the taller pole be $H_T = 8$ m and its top be $T$. Let its foot be $A$.
Let the height of the smaller pole be $H_R = 7$ m and its top be $R$. Let its foot be $C$.
Point $P$ is on the ground. The smaller pole is between the taller pole and point $P$.
The distance from point $P$ to the smaller pole is $PC = 20$ m.

We can visualize this with a diagram:
Let $P$ be the origin $(0,0)$ on the ground.
The foot of the smaller pole $C$ is at $(20,0)$. The top of the smaller pole $R$ is at $(20,7)$.
Let the horizontal distance between the feet of the poles be $AC = x$.
The foot of the taller pole $A$ is at $(20+x, 0)$. The top of the taller pole $T$ is at $(20+x, 8)$.

Now, consider the triangle formed by $R$, $T$, and a point $E$ directly below $T$ on the horizontal line passing through $R$.
The coordinates of $R$ are $(20,7)$.
The coordinates of $T$ are $(20+x, 8)$.
The coordinates of $E$ are $(20+x, 7)$.
This forms a right-angled triangle $\triangle RET$ with the right angle at $E$.
The horizontal distance $RE = (20+x) - 20 = x$.
The vertical distance $ET = 8 - 7 = 1$ m.
The angle of elevation of $T$ from $R$ is $\angle TRE = 30^\circ$.

a) The distance RT.

Step 1: Use the sine function in $\triangle RET$.

$$\sin(\angle TRE) = \frac{ET}{RT}$$ $$\sin(30^\circ) = \frac{1}{RT}$$

Step 2: Solve for $RT$.

$$RT = \frac{1}{\sin(30^\circ)}$$ $$RT = \frac{1}{0.5}$$ $$RT = 2 \text{ m}$$

The distance $RT$ is $\boxed{2 \text{ m}}$.

b) The distance of the foot of the taller pole from P (PA).

Step 1: Find the horizontal distance $x$ (which is $AC$) using the tangent function in $\triangle RET$.

$$\tan(\angle TRE) = \frac{ET}{RE}$$ $$\tan(30^\circ) = \frac{1}{x}$$

Step 2: Solve for $x$.

$$x = \frac{1}{\tan(30^\circ)}$$ $$x = \frac{1}{1/\sqrt{3}}$$ $$x = \sqrt{3} \text{ m}$$

Step 3: Calculate the distance $PA$.
The distance $PA = PC + AC = 20 + x$.

$$PA = 20 + \sqrt{3} \text{ m}$$ $$PA \approx 20 + 1.7320508$$ $$PA \approx 21.7320508 \text{ m}$$

Step 4: Round to 3 significant figures.

$$PA \approx 21.7 \text{ m}$$

The distance $PA$ is $\boxed{21.7 \text{ m}}$.

c) The angle of elevation of T from P, correct to one decimal point.

Step 1: Consider the right-angled triangle $\triangle PAT$.
The height of the taller pole is $AT = 8$ m.
The distance $PA = 20 + \sqrt{3}$ m.

Step 2: Use the tangent function to find $\angle TPA$.

$$\tan(\angle TPA) = \frac{AT}{PA}$$ $$\tan(\angle TPA) = \frac{8}{20 + \sqrt{3}}$$ $$\tan(\angle TPA) \approx \frac{8}{21.7320508}$$ $$$$