AB is a vertical pole with B at the ground level and and A at the top .A man finds that the angle of elevation of the point A from a certain point C on the ground is `60^(@)` . He moves away from the pole along the line BC to a point D such that CD = 7 m .From D . the angle of elevation of the point A is `45^(@)` .Find the height of the pole .

Options

A. `(7sqrt(3))/(2)(1)/(sqrt(3)+1)m`
B. `(7sqrt(3))/(2)(1)/(sqrt(3)-1)m`
C. `(7sqrt(3))/(2)(sqrt(3)+1)m`
D. `(7sqrt(3))/(2)(sqrt(3)-1)m`

Answer

To find the height of the pole AB, we will follow these steps:

### Step 1: Set Up the Problem
Let the height of the pole AB be \( H \). The point C is where the angle of elevation to point A is \( 60^\circ \), and the point D is where the angle of elevation to point A is \( 45^\circ \). The distance CD is given as \( 7 \) m.

### Step 2: Use Triangle ABC
From point C, we can use the tangent of the angle of elevation:
\[ \tan(60^\circ) = \frac{H}{BC} \]
Since \( \tan(60^\circ) = \sqrt{3} \), we have:
\[ \sqrt{3} = \frac{H}{BC} \]
This gives us our first equation:
\[ H = BC \cdot \sqrt{3} \quad \text{(1)} \]

### Step 3: Use Triangle ABD
From point D, we can use the tangent of the angle of elevation:
\[ \tan(45^\circ) = \frac{H}{BD} \]
Since \( \tan(45^\circ) = 1 \), we have:
\[ 1 = \frac{H}{BD} \]
This gives us:
\[ H = BD \quad \text{(2)} \]
Now, we know that \( BD = BC + CD \), and since \( CD = 7 \) m, we can write:
\[ BD = BC + 7 \quad \text{(3)} \]

### Step 4: Substitute Equation (3) into Equation (2)
From equation (2):
\[ H = BC + 7 \]
Now we can substitute equation (1) into this:
\[ BC \cdot \sqrt{3} = BC + 7 \]

### Step 5: Rearrange the Equation
Rearranging gives:
\[ BC \cdot \sqrt{3} - BC = 7 \]
Factoring out \( BC \):
\[ BC(\sqrt{3} - 1) = 7 \]
Thus, we can solve for \( BC \):
\[ BC = \frac{7}{\sqrt{3} - 1} \quad \text{(4)} \]

### Step 6: Substitute Back to Find H
Now substitute equation (4) back into equation (1):
\[ H = \left(\frac{7}{\sqrt{3} - 1}\right) \cdot \sqrt{3} \]
This simplifies to:
\[ H = \frac{7\sqrt{3}}{\sqrt{3} - 1} \]

### Step 7: Rationalize the Denominator
To rationalize the denominator:
\[ H = \frac{7\sqrt{3}(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)} = \frac{7\sqrt{3}(\sqrt{3} + 1)}{3 - 1} = \frac{7\sqrt{3}(\sqrt{3} + 1)}{2} \]

### Final Answer
Thus, the height of the pole \( H \) is:
\[ H = \frac{7\sqrt{3}(\sqrt{3} + 1)}{2} \]

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