The shadow of a tower, when the angle of elevation of the sun is `45^@` , is found to be 15 metres longer than when it is `60^@` . Find the height of the tower. (Use `sqrt3 = 1.732` )

Options

A. 38.92 m
B. 36.59 m
C. 35.49 m
D. 37.49 m

Answer

To find the height of the tower based on the given conditions, we can follow these steps:

### Step 1: Set up the problem
Let the height of the tower be \( H \) meters. When the angle of elevation of the sun is \( 45^\circ \), let the length of the shadow be \( x \) meters. When the angle of elevation is \( 60^\circ \), the shadow is \( x - 15 \) meters (since it's 15 meters shorter).

### Step 2: Use trigonometric ratios
For the angle of elevation \( 60^\circ \):
\[ \tan(60^\circ) = \frac{H}{x - 15} \]
Using the value of \( \tan(60^\circ) = \sqrt{3} \):
\[ \sqrt{3} = \frac{H}{x - 15} \]
From this, we can express \( H \) in terms of \( x \):
\[ H = \sqrt{3}(x - 15) \quad \text{(Equation 1)} \]

For the angle of elevation \( 45^\circ \):
\[ \tan(45^\circ) = \frac{H}{x} \]
Since \( \tan(45^\circ) = 1 \):
\[ 1 = \frac{H}{x} \]
Thus, we can express \( H \) as:
\[ H = x \quad \text{(Equation 2)} \]

### Step 3: Set the equations equal to each other
Now we have two expressions for \( H \):
\[ \sqrt{3}(x - 15) = x \]

### Step 4: Solve for \( x \)
Rearranging the equation:
\[ \sqrt{3}x - 15\sqrt{3} = x \]
Bringing all terms involving \( x \) to one side:
\[ \sqrt{3}x - x = 15\sqrt{3} \]
Factoring out \( x \):
\[ x(\sqrt{3} - 1) = 15\sqrt{3} \]
Now, solving for \( x \):
\[ x = \frac{15\sqrt{3}}{\sqrt{3} - 1} \]

### Step 5: Rationalize the denominator
To rationalize the denominator:
\[ x = \frac{15\sqrt{3}(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)} = \frac{15\sqrt{3}(\sqrt{3} + 1)}{3 - 1} = \frac{15\sqrt{3}(\sqrt{3} + 1)}{2} \]
Calculating the numerator:
\[ = \frac{15(3 + \sqrt{3})}{2} = \frac{45 + 15\sqrt{3}}{2} \]

### Step 6: Substitute back to find \( H \)
Now substitute \( x \) back into Equation 2 to find \( H \):
\[ H = x = \frac{45 + 15\sqrt{3}}{2} \]

### Step 7: Calculate \( H \) using \( \sqrt{3} = 1.732 \)
Substituting the value of \( \sqrt{3} \):
\[ H = \frac{45 + 15 \times 1.732}{2} = \frac{45 + 25.98}{2} = \frac{70.98}{2} = 35.49 \]

### Conclusion
Thus, the height of the tower is \( 35.49 \) meters.

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