A tower `T_(1)` of height 60 m is located exactly opposite to a tower `T_(2)` of height 80 m on a straight road. From the top of `T_(1)`, if the angle of depression of the foot of `T_(2)` is twice the angle of elevation of the top of `T_(2)`, then the width (in m) of the road between the feet of the towers `T_(1)` and `T_(2)` is

Options

A. `20sqrt(3)`
B. `10sqrt(3)`
C. `10sqrt(2)`
D. `20sqrt(2)`

Answer

To solve the problem, we will follow these steps:

### Step 1: Understanding the Problem
We have two towers, \( T_1 \) and \( T_2 \), with heights of 60 m and 80 m respectively. The angle of depression from the top of tower \( T_1 \) to the foot of tower \( T_2 \) is twice the angle of elevation from the foot of tower \( T_2 \) to the top of tower \( T_2 \).

### Step 2: Setting Up the Diagram
Let's denote:
- \( A \): the top of tower \( T_1 \)
- \( B \): the foot of tower \( T_1 \)
- \( C \): the top of tower \( T_2 \)
- \( D \): the foot of tower \( T_2 \)
- \( W \): the width of the road between \( B \) and \( D \)

### Step 3: Defining Angles
Let:
- \( \theta \): the angle of elevation from \( D \) to \( C \)
- The angle of depression from \( A \) to \( D \) is \( 2\theta \).

### Step 4: Using Trigonometric Ratios
From triangle \( ABD \):
- The height of tower \( T_1 \) is 60 m.
- The angle of depression \( \angle ADB = 2\theta \).

Using the tangent function:
\[ \tan(2\theta) = \frac{60}{W} \]

From triangle \( DBC \):
- The height of tower \( T_2 \) is 80 m.
- The angle of elevation \( \angle DBC = \theta \).

Using the tangent function:
\[ \tan(\theta) = \frac{80 - 60}{W} = \frac{20}{W} \]

### Step 5: Using the Double Angle Formula
We know that:
\[ \tan(2\theta) = \frac{2\tan(\theta)}{1 - \tan^2(\theta)} \]

Substituting \( \tan(\theta) = \frac{20}{W} \):
\[ \tan(2\theta) = \frac{2 \cdot \frac{20}{W}}{1 - \left(\frac{20}{W}\right)^2} \]

### Step 6: Equating the Two Expressions for \( \tan(2\theta) \)
Now we equate the two expressions for \( \tan(2\theta) \):
\[ \frac{60}{W} = \frac{40/W}{1 - \frac{400}{W^2}} \]

### Step 7: Cross Multiplying
Cross multiplying gives:
\[ 60(1 - \frac{400}{W^2}) = 40 \]

### Step 8: Simplifying the Equation
Expanding and simplifying:
\[ 60 - \frac{24000}{W^2} = 40 \]
\[ 20 = \frac{24000}{W^2} \]
\[ W^2 = \frac{24000}{20} = 1200 \]

### Step 9: Finding \( W \)
Taking the square root:
\[ W = \sqrt{1200} = 20\sqrt{3} \]

### Final Answer
The width of the road between the feet of the towers \( T_1 \) and \( T_2 \) is \( 20\sqrt{3} \) meters.

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