Two poles of height 10 meters and 20 meters stand at the centres of two circular plots which touch each other externally at a point and the two poles subtend angles `30^(@) and 60^(@)` respectively at this point, then the distance between the centres of these circular plots is

Options

A. 30 meters
B. `(50)/(sqrt3)` meters
C. `(70)/(sqrt3)` meters
D. `(10sqrt3+20)` meters

Answer

To solve the problem step by step, we will use trigonometric relationships in right triangles formed by the poles and the points where the circles touch.

### Step 1: Understand the Geometry
We have two poles:
- Pole A (height = 20 m)
- Pole B (height = 10 m)

These poles are at the centers of two circular plots that touch each other externally at point C.

### Step 2: Identify the Angles
The angles subtended by the poles at point C are:
- Angle subtended by pole A (20 m) = 60°
- Angle subtended by pole B (10 m) = 30°

### Step 3: Draw the Right Triangles
From point C, we can form two right triangles:
- Triangle ADC (for pole A)
- Triangle BEC (for pole B)

### Step 4: Use Trigonometric Ratios
For triangle ADC:
- The height (perpendicular) = AD = 20 m
- Angle ACD = 60°
- We need to find AC (the base).

Using the cotangent function:
\[ \cot(60°) = \frac{\text{Base}}{\text{Height}} \implies AC = AD \cdot \cot(60°) \]
\[ \cot(60°) = \frac{1}{\sqrt{3}} \implies AC = 20 \cdot \frac{1}{\sqrt{3}} = \frac{20}{\sqrt{3}} \]

For triangle BEC:
- The height (perpendicular) = BE = 10 m
- Angle BCE = 30°
- We need to find CB (the base).

Using the cotangent function:
\[ \cot(30°) = \frac{\text{Base}}{\text{Height}} \implies CB = BE \cdot \cot(30°) \]
\[ \cot(30°) = \sqrt{3} \implies CB = 10 \cdot \sqrt{3} = 10\sqrt{3} \]

### Step 5: Calculate the Total Distance AB
The total distance between the centers of the circular plots (AB) is the sum of AC and CB:
\[ AB = AC + CB = \frac{20}{\sqrt{3}} + 10\sqrt{3} \]

### Step 6: Simplify the Expression
To combine these fractions, we can express \(10\sqrt{3}\) with a common denominator:
\[ 10\sqrt{3} = \frac{10\sqrt{3} \cdot \sqrt{3}}{\sqrt{3}} = \frac{30}{\sqrt{3}} \]
Now, we can add:
\[ AB = \frac{20}{\sqrt{3}} + \frac{30}{\sqrt{3}} = \frac{50}{\sqrt{3}} \]

### Final Answer
Thus, the distance between the centers of the circular plots is:
\[ AB = \frac{50}{\sqrt{3}} \text{ meters} \]