If the angle of elevation of the Sun changes from `30^(@)` to `45^(@)` , the length of the shadow of a piller decreases by 20 metres .The height of the pillar is

Options

A. `20(sqrt(3)-1)m`
B. `20(sqrt(3)+1)m`
C. `10(sqrt(3)-1)`m
D. `10(sqrt(3)+1)` m

Answer

To solve the problem, we will use trigonometric ratios and the properties of right triangles. Let's break it down step by step.

### Step 1: Understanding the Problem
We have a pillar of height \( h \) and two angles of elevation of the sun: \( 30^\circ \) and \( 45^\circ \). The length of the shadow decreases by 20 meters when the angle changes from \( 30^\circ \) to \( 45^\circ \).

### Step 2: Setting Up the Triangles
1. **For \( 30^\circ \)**:
- Let the length of the shadow be \( x \).
- Using the tangent function:
\[ \tan(30^\circ) = \frac{h}{x} \]
- We know that \( \tan(30^\circ) = \frac{1}{\sqrt{3}} \), so:
\[ \frac{1}{\sqrt{3}} = \frac{h}{x} \implies h = \frac{x}{\sqrt{3}} \]

2. **For \( 45^\circ \)**:
- The length of the shadow is now \( x - 20 \) meters.
- Again using the tangent function:
\[ \tan(45^\circ) = \frac{h}{x - 20} \]
- We know that \( \tan(45^\circ) = 1 \), so:
\[ 1 = \frac{h}{x - 20} \implies h = x - 20 \]

### Step 3: Setting Up the Equation
Now we have two expressions for \( h \):
1. From the first triangle: \( h = \frac{x}{\sqrt{3}} \)
2. From the second triangle: \( h = x - 20 \)

Setting these equal to each other:
\[ \frac{x}{\sqrt{3}} = x - 20 \]

### Step 4: Solving for \( x \)
To solve for \( x \), we can rearrange the equation:
\[ x - \frac{x}{\sqrt{3}} = 20 \]
Factoring out \( x \):
\[ x \left(1 - \frac{1}{\sqrt{3}}\right) = 20 \]
Calculating \( 1 - \frac{1}{\sqrt{3}} \):
\[ 1 - \frac{1}{\sqrt{3}} = \frac{\sqrt{3} - 1}{\sqrt{3}} \]
Thus:
\[ x \cdot \frac{\sqrt{3} - 1}{\sqrt{3}} = 20 \]
Multiplying both sides by \( \frac{\sqrt{3}}{\sqrt{3} - 1} \):
\[ x = \frac{20\sqrt{3}}{\sqrt{3} - 1} \]

### Step 5: Finding \( h \)
Now substitute \( x \) back into one of the expressions for \( h \):
Using \( h = x - 20 \):
\[ h = \frac{20\sqrt{3}}{\sqrt{3} - 1} - 20 \]
To simplify, we can express 20 with a common denominator:
\[ h = \frac{20\sqrt{3}}{\sqrt{3} - 1} - \frac{20(\sqrt{3} - 1)}{\sqrt{3} - 1} \]
\[ h = \frac{20\sqrt{3} - 20(\sqrt{3} - 1)}{\sqrt{3} - 1} \]
\[ h = \frac{20\sqrt{3} - 20\sqrt{3} + 20}{\sqrt{3} - 1} = \frac{20}{\sqrt{3} - 1} \]

### Step 6: Rationalizing the Denominator
To rationalize:
\[ h = \frac{20(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)} = \frac{20(\sqrt{3} + 1)}{3 - 1} = \frac{20(\sqrt{3} + 1)}{2} = 10(\sqrt{3} + 1) \]

### Final Answer
Thus, the height of the pillar is:
\[ h = 10(\sqrt{3} + 1) \text{ meters} \]