The electric pole subtends an angle of `30^(@)` at a point on the same level as its foot. At a second point 'b' metres above the first, the depression of the foot of the tower is `60^(@)`. The height of the tower (in towers) is equal to

Options

A. b m
B. 3b m
C. `sqrt(3)`b m
D. `(b//3)`m

Answer

To solve the problem, we need to find the height of the electric pole (tower) based on the angles given and the height difference between two points. Let's break down the solution step by step.

### Step 1: Understand the problem
We have an electric pole (tower) of height \( H \). At point \( D \), which is at the foot of the tower, the angle of elevation to the top of the tower (point \( A \)) is \( 30^\circ \). At point \( C \), which is \( b \) meters above point \( D \), the angle of depression to point \( D \) is \( 60^\circ \).

### Step 2: Set up the triangles
1. **Triangle \( BEC \)**: Here, \( BE \) is the height difference \( b \) (the height from point \( D \) to point \( C \)), and \( EC \) is the horizontal distance from point \( D \) to the foot of the tower.
2. **Triangle \( ABD \)**: In this triangle, \( AB \) is the height of the tower \( H \), and \( BD \) is the same horizontal distance \( EC \).

### Step 3: Use trigonometric ratios
1. For triangle \( BEC \):
\[ \tan(60^\circ) = \frac{BE}{EC} \]
Since \( BE = b \) and \( \tan(60^\circ) = \sqrt{3} \), we can write:
\[ \sqrt{3} = \frac{b}{EC} \]
Rearranging gives:
\[ EC = \frac{b}{\sqrt{3}} \quad \text{(Equation 1)} \]

2. For triangle \( ABD \):
\[ \tan(30^\circ) = \frac{AB}{BD} \]
Here, \( AB = H \) and \( BD = EC \). Since \( \tan(30^\circ) = \frac{1}{\sqrt{3}} \), we have:
\[ \frac{1}{\sqrt{3}} = \frac{H}{EC} \]
Rearranging gives:
\[ H = EC \cdot \frac{1}{\sqrt{3}} \quad \text{(Equation 2)} \]

### Step 4: Substitute Equation 1 into Equation 2
From Equation 1, we know:
\[ EC = \frac{b}{\sqrt{3}} \]
Substituting this into Equation 2:
\[ H = \left(\frac{b}{\sqrt{3}}\right) \cdot \frac{1}{\sqrt{3}} = \frac{b}{3} \]

### Conclusion
The height of the tower \( H \) is:
\[ H = \frac{b}{3} \text{ meters} \]