A man standing in one corner of a square football field observes that the angle subtended by a pole in the corner just diagonally opposite to this corner is `60^@`. When he retires 80 m from the corner, along the same straight line, he finds the angle to be `30^@`. The length of the field, in m, is :

Options

A. 40
B. `20 sqrt2`
C. 20
D. `40 sqrt 2`

Answer

To solve the problem step by step, we will analyze the situation using trigonometry and geometry.

### Step 1: Understand the Setup
We have a square football field, and we denote the corners as A, B, C, and D. The man is standing at corner A, and the pole is at corner C, which is diagonally opposite to A. The man observes an angle of 60° when looking at the pole.

### Step 2: Set Up the Diagram
Let the length of each side of the square field be \( a \). The distance from point A to point C (the diagonal) can be represented as \( AC = a\sqrt{2} \).

### Step 3: Use the First Angle (60°)
From point A, the angle subtended by the pole at C is 60°. We can use the tangent function to express the height of the pole (let's denote it as \( h \)) in terms of \( a \):
\[ \tan(60^\circ) = \frac{h}{a} \]
Since \( \tan(60^\circ) = \sqrt{3} \), we have:
\[ \sqrt{3} = \frac{h}{a} \implies h = a\sqrt{3} \]

### Step 4: Move 80 m Back and Use the Second Angle (30°)
The man then moves back 80 m along the same line. Now he is at point K, where \( AK = a + 80 \). The angle subtended by the pole at C is now 30°:
\[ \tan(30^\circ) = \frac{h}{a + 80} \]
Since \( \tan(30^\circ) = \frac{1}{\sqrt{3}} \), we have:
\[ \frac{1}{\sqrt{3}} = \frac{h}{a + 80} \implies h = \frac{(a + 80)}{\sqrt{3}} \]

### Step 5: Set the Two Expressions for Height Equal
Now we have two expressions for \( h \):
1. \( h = a\sqrt{3} \)
2. \( h = \frac{(a + 80)}{\sqrt{3}} \)

Setting them equal to each other:
\[ a\sqrt{3} = \frac{(a + 80)}{\sqrt{3}} \]

### Step 6: Cross Multiply and Solve for \( a \)
Cross multiplying gives:
\[ 3a^2 = a + 80 \]
Rearranging this gives:
\[ 3a^2 - a - 80 = 0 \]

### Step 7: Use the Quadratic Formula
To solve for \( a \), we use the quadratic formula:
\[ a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 3, b = -1, c = -80 \):
\[ a = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 3 \cdot (-80)}}{2 \cdot 3} \]
\[ = \frac{1 \pm \sqrt{1 + 960}}{6} \]
\[ = \frac{1 \pm \sqrt{961}}{6} \]
\[ = \frac{1 \pm 31}{6} \]

### Step 8: Calculate the Values
Calculating the two possible values:
1. \( a = \frac{32}{6} = \frac{16}{3} \) (not valid since it must be positive)
2. \( a = \frac{-30}{6} = -5 \) (not valid)

Thus, we only consider the positive root:
\[ a = \frac{32}{6} = \frac{16}{3} \text{ (valid)} \]

### Step 9: Find the Length of the Field
The length of the field is \( a \), which is \( \frac{16}{3} \) meters. To express this in a more standard form, we can convert it to decimal or leave it as a fraction.

### Final Answer
The length of the field is \( \frac{16}{3} \) meters or approximately \( 5.33 \) meters.