When the elevation of the sun changes from `45^(@) "to " 30^(@)`, the shadow of a tower increases by 60 units then the height of the tower is

Options

A. `30sqrt(3)` units
B. `30sqrt(2)` units
C. `30(sqrt(3)+1)` units
D. `30(sqrt(2)+1)` units

Answer

To solve the problem, we need to find the height of the tower based on the change in the shadow length as the elevation of the sun changes from 45 degrees to 30 degrees.

Let's denote the height of the tower as \( H \) and the length of the shadow when the sun is at 45 degrees as \( B \).

### Step 1: Set up the equations using trigonometry

1. **When the sun is at 45 degrees:**
\[ \tan(45^\circ) = \frac{H}{B} \]
Since \( \tan(45^\circ) = 1 \), we have:
\[ H = B \]

2. **When the sun is at 30 degrees:**
The length of the shadow increases by 60 units, so the new length of the shadow is \( B + 60 \).
\[ \tan(30^\circ) = \frac{H}{B + 60} \]
Since \( \tan(30^\circ) = \frac{1}{\sqrt{3}} \), we have:
\[ \frac{H}{B + 60} = \frac{1}{\sqrt{3}} \]
Rearranging gives:
\[ H = \frac{1}{\sqrt{3}}(B + 60) \]

### Step 2: Substitute \( B \) from the first equation into the second equation

From the first equation, we know \( B = H \). Substituting this into the second equation:
\[ H = \frac{1}{\sqrt{3}}(H + 60) \]

### Step 3: Solve for \( H \)

Multiply both sides by \( \sqrt{3} \) to eliminate the fraction:
\[ \sqrt{3}H = H + 60 \]
Rearranging gives:
\[ \sqrt{3}H - H = 60 \]
\[ H(\sqrt{3} - 1) = 60 \]
Now, divide both sides by \( \sqrt{3} - 1 \):
\[ H = \frac{60}{\sqrt{3} - 1} \]

### Step 4: Rationalize the denominator

To rationalize the denominator, multiply the numerator and denominator by \( \sqrt{3} + 1 \):
\[ H = \frac{60(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)} = \frac{60(\sqrt{3} + 1)}{3 - 1} = \frac{60(\sqrt{3} + 1)}{2} = 30(\sqrt{3} + 1) \]

### Final Answer

Thus, the height of the tower is:
\[ H = 30\sqrt{3} + 30 \]