From the top of a pillar of height 20 m, theangles of elevation and depression of the top and bottom of another pillar are `30^(@)` and `45^(@)`, respectively. Then, the height of the second pillar (in m) is

Options

A. `(20)/(sqrt3)(sqrt3-1)`
B. `(20)/(sqrt3)(sqrt3+1)`
C. `20sqrt3`
D. `(20)/(sqrt3)`

Answer

To solve the problem step by step, we will use trigonometric ratios and the information given in the question.

### Step-by-Step Solution:

1. **Understand the Problem**: We have two pillars. The first pillar (AB) has a height of 20 m. From the top of this pillar, the angle of depression to the bottom of the second pillar (CD) is 45°, and the angle of elevation to the top of the second pillar (EF) is 30°.

2. **Draw the Diagram**:
- Let A be the top of the first pillar, B be the bottom of the first pillar (ground level), C be the bottom of the second pillar, and D be the top of the second pillar.
- The height of the first pillar AB = 20 m.
- The angle of depression from A to C is 45°.
- The angle of elevation from A to D is 30°.

3. **Identify Distances**:
- Let the distance from the base of the first pillar (B) to the base of the second pillar (C) be D.
- Let the height of the second pillar (CD) be h.

4. **Using Triangle ACB (for angle of depression)**:
- In triangle ACB, angle ACB = 45°.
- Using the tangent function:
\[ \tan(45°) = \frac{AB}{BC} = \frac{20}{D} \]
- Since \(\tan(45°) = 1\), we have:
\[ 1 = \frac{20}{D} \implies D = 20 \text{ m} \]

5. **Using Triangle ADB (for angle of elevation)**:
- In triangle ADB, angle ADB = 30°.
- Again using the tangent function:
\[ \tan(30°) = \frac{AD}{AB} = \frac{h}{D} \]
- We know that \(\tan(30°) = \frac{1}{\sqrt{3}}\), so:
\[ \frac{1}{\sqrt{3}} = \frac{h}{20} \]
- Rearranging gives:
\[ h = \frac{20}{\sqrt{3}} \text{ m} \]

6. **Finding the Total Height of the Second Pillar**:
- The total height of the second pillar (CD) is given by:
\[ \text{Total Height} = CD = h + BC \]
- We already know \(BC = 20\) m and \(h = \frac{20}{\sqrt{3}}\) m.
- Therefore:
\[ CD = 20 + \frac{20}{\sqrt{3}} = 20\left(1 + \frac{1}{\sqrt{3}}\right) \]
- Simplifying gives:
\[ CD = 20\left(\frac{\sqrt{3} + 1}{\sqrt{3}}\right) = \frac{20(\sqrt{3} + 1)}{\sqrt{3}} \text{ m} \]

### Final Answer:
The height of the second pillar is \( \frac{20(\sqrt{3} + 1)}{\sqrt{3}} \) m.