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 find the height of the tower when the elevation of the sun changes from \( 45^\circ \) to \( 30^\circ \) and the shadow increases by 60 units, we can follow these steps:

### Step 1: Define Variables
Let the height of the tower be \( h \) (which we need to find).

### Step 2: Analyze the Situation
When the sun is at an elevation of \( 45^\circ \):
- The length of the shadow can be calculated using the tangent function:
\[ \tan(45^\circ) = \frac{h}{s_1} \implies s_1 = h \]
(since \( \tan(45^\circ) = 1 \))

When the sun is at an elevation of \( 30^\circ \):
- The length of the shadow changes to:
\[ \tan(30^\circ) = \frac{h}{s_2} \implies s_2 = h \cdot \sqrt{3} \]
(since \( \tan(30^\circ) = \frac{1}{\sqrt{3}} \))

### Step 3: Set Up the Equation
According to the problem, the increase in shadow length is 60 units:
\[ s_2 - s_1 = 60 \]
Substituting the values of \( s_1 \) and \( s_2 \):
\[ h \cdot \sqrt{3} - h = 60 \]

### Step 4: Factor Out \( h \)
\[ h(\sqrt{3} - 1) = 60 \]

### Step 5: Solve for \( h \)
\[ h = \frac{60}{\sqrt{3} - 1} \]

### Step 6: Rationalize the Denominator
To rationalize the denominator, multiply the numerator and denominator by \( \sqrt{3} + 1 \):
\[ h = \frac{60(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)} = \frac{60(\sqrt{3} + 1)}{3 - 1} = \frac{60(\sqrt{3} + 1)}{2} \]
\[ h = 30(\sqrt{3} + 1) \]

### Conclusion
Thus, the height of the tower is:
\[ h = 30(\sqrt{3} + 1) \]