Two pillars of equal heights stand on either side of a roadway, which is 150 m wide. At a point in the roadway between the pillars the elevations of the tops of the pillars are `60^(@) and 30^(@)` . Find the height of the pillars and the position of the point

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the Problem
We have two pillars (AB and CD) of equal height (H) on either side of a roadway that is 150 meters wide. The angles of elevation from a point E in the roadway to the tops of the pillars are 60 degrees for pillar CD and 30 degrees for pillar AB.

### Step 2: Set Up the Diagram
Let's denote:
- AB as the first pillar on the left.
- CD as the second pillar on the right.
- The distance between the two pillars (BD) is 150 meters.
- The height of both pillars is H.
- The distance from point E to pillar AB is BE, and the distance from point E to pillar CD is CE.

### Step 3: Use Trigonometry for Pillar CD
From point E, the angle of elevation to the top of pillar CD is 60 degrees. Using the tangent function:
\[ \tan(60^\circ) = \frac{H}{CE} \]
Since \(\tan(60^\circ) = \sqrt{3}\), we can write:
\[ \sqrt{3} = \frac{H}{CE} \implies CE = \frac{H}{\sqrt{3}} \quad \text{(Equation 1)} \]

### Step 4: Use Trigonometry for Pillar AB
From point E, the angle of elevation to the top of pillar AB is 30 degrees. Using the tangent function:
\[ \tan(30^\circ) = \frac{H}{BE} \]
Since \(\tan(30^\circ) = \frac{1}{\sqrt{3}}\), we can write:
\[ \frac{1}{\sqrt{3}} = \frac{H}{BE} \implies BE = H \sqrt{3} \quad \text{(Equation 2)} \]

### Step 5: Relate Distances
We know that the total distance between the two pillars is:
\[ BD = BE + CE = 150 \text{ meters} \]
Substituting the values from Equations 1 and 2:
\[ 150 = H \sqrt{3} + \frac{H}{\sqrt{3}} \]

### Step 6: Solve for H
To combine the terms, we can express them with a common denominator:
\[ 150 = H \sqrt{3} + \frac{H}{\sqrt{3}} = H \left(\sqrt{3} + \frac{1}{\sqrt{3}}\right) \]
\[ = H \left(\frac{3 + 1}{\sqrt{3}}\right) = H \left(\frac{4}{\sqrt{3}}\right) \]
Now, we can solve for H:
\[ 150 = H \left(\frac{4}{\sqrt{3}}\right) \]
\[ H = \frac{150 \sqrt{3}}{4} = 37.5 \sqrt{3} \]
Calculating \(H\):
\[ H \approx 64.95 \text{ meters} \]

### Step 7: Find the Position of Point E
Using Equation 1 to find CE:
\[ CE = \frac{H}{\sqrt{3}} = \frac{64.95}{\sqrt{3}} \approx 37.5 \text{ meters} \]

### Final Results
- Height of the pillars (H) is approximately **64.95 meters**.
- The distance from point E to pillar AB (BE) is approximately **37.5 meters**.