If the sun ray' inclination increases from `45^(@)` to `60^(@)` the length of the shadow of a tower decreases by 50 m. Find the height of the tower (in m).

Options

A. `50(sqrt3-1)`
B. `75(3-sqrt3)`
C. `100(sqrt3+1)`
D. `25(3+sqrt3)`

Explanation

To solve the problem, we will use the concept of trigonometry, specifically the tangent function, which relates the angle of elevation of the sun to the height of the tower and the length of the shadow.

**Step 1: Define the variables.**
Let:
- \( h \) = height of the tower (in meters)
- \( L_1 \) = length of the shadow when the sun's inclination is \( 45^\circ \)
- \( L_2 \) = length of the shadow when the sun's inclination is \( 60^\circ \)

According to the problem, we know that:
\[ L_1 - L_2 = 50 \, \text{m} \]

**Step 2: Use the tangent function for both angles.**
From trigonometry, we know that:
- For \( 45^\circ \):
\[ \tan(45^\circ) = \frac{h}{L_1} \implies L_1 = h \]
- For \( 60^\circ \):
\[ \tan(60^\circ) = \frac{h}{L_2} \implies L_2 = \frac{h}{\sqrt{3}} \]

**Step 3: Substitute the expressions for \( L_1 \) and \( L_2 \) into the equation.**
Substituting \( L_1 \) and \( L_2 \) into the equation \( L_1 - L_2 = 50 \):
\[ h - \frac{h}{\sqrt{3}} = 50 \]

**Step 4: Solve for \( h \).**
To solve for \( h \), first factor out \( h \):
\[ h \left(1 - \frac{1}{\sqrt{3}}\right) = 50 \]
Now, simplify the expression in the parentheses:
\[ 1 - \frac{1}{\sqrt{3}} = \frac{\sqrt{3} - 1}{\sqrt{3}} \]
So we have:
\[ h \cdot \frac{\sqrt{3} - 1}{\sqrt{3}} = 50 \]
Now, multiply both sides by \( \frac{\sqrt{3}}{\sqrt{3} - 1} \):
\[ h = 50 \cdot \frac{\sqrt{3}}{\sqrt{3} - 1} \]

**Step 5: Calculate the value of \( h \).**
Using the approximate value of \( \sqrt{3} \approx 1.732 \):
\[ h = 50 \cdot \frac{1.732}{1.732 - 1} = 50 \cdot \frac{1.732}{0.732} \approx 50 \cdot 2.365 = 118.25 \, \text{m} \]

**Final Answer:**
The height of the tower is approximately \( 118.25 \, \text{m} \).

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