The two banks of a canal are straight and parallel .A,B,C are three persons of whom A stands on one bank and B and C on the opposite banks ,B finds the angle ABC is `30^(@)`,while C finds that the angle ACB `60^(@)` .If B and C are 100 metres apart ,the breadth of the canal is

Options

A. `(25)/(sqrt(3))` metres
B. `20sqrt(3)` metres
C. `25sqrt(3)` metres
D. `(20)/(sqrt(3))` metres

Answer

To solve the problem, we will use the properties of triangles and trigonometric ratios. Here's a step-by-step solution:

### Step 1: Understand the Setup
We have a canal with two banks that are parallel. Points A, B, and C represent three persons, where A is on one bank and B and C are on the opposite bank. The angles given are:
- Angle ABC = 30°
- Angle ACB = 60°

### Step 2: Identify the Triangle
From the information given, we can form triangle ABC. The sides of the triangle are:
- BC = 100 meters (the distance between B and C)
- We need to find the height (the breadth of the canal), which is the perpendicular distance from point A to line BC.

### Step 3: Use the Angles to Find the Height
In triangle ABC:
- The angle at B (angle ABC) is 30°.
- The angle at C (angle ACB) is 60°.
- The angle at A (angle BAC) can be calculated as:
\[ \text{Angle A} = 180° - (30° + 60°) = 90° \]
This confirms that triangle ABC is a right triangle with angle A being the right angle.

### Step 4: Apply Trigonometric Ratios
In a right triangle, we can use the tangent function to find the height (breadth of the canal). The tangent of an angle in a right triangle is the ratio of the opposite side to the adjacent side.

1. For angle ABC (30°):
\[ \tan(30°) = \frac{\text{height}}{BC} \]
Here, BC = 100 meters.
\[ \tan(30°) = \frac{\text{height}}{100} \]
We know that \(\tan(30°) = \frac{1}{\sqrt{3}}\):
\[ \frac{1}{\sqrt{3}} = \frac{\text{height}}{100} \]
Thus, the height (h) is:
\[ h = \frac{100}{\sqrt{3}} \approx 57.74 \text{ meters} \]

2. For angle ACB (60°):
\[ \tan(60°) = \frac{\text{height}}{AB} \]
Here, we can use the relationship between the sides:
\[ \tan(60°) = \sqrt{3} = \frac{\text{height}}{AB} \]
Rearranging gives:
\[ \text{height} = AB \cdot \sqrt{3} \]

### Step 5: Solve for the Height
Using the 30-60-90 triangle properties:
- The side opposite the 30° angle (height) is half the hypotenuse (BC).
- The side opposite the 60° angle is \(\frac{\sqrt{3}}{2}\) times the hypotenuse.

Thus, we can calculate the height (breadth of the canal):
\[ \text{height} = 100 \cdot \frac{1}{2} \cdot \sqrt{3} = 50\sqrt{3} \]

### Final Answer
The breadth of the canal is:
\[ \text{Breadth} = 50\sqrt{3} \text{ meters} \]