The angle of elevation of the top of an unfinished tower at a distance of 100 m from its base is `45^(@)`. How much higher must the tower be raised so that angle of elevation of its top at the same point may be `60^(@)`?

Options

A. `50sqrt(2)` m
B. 100 m
C. `100(sqrt(3)-1)` m
D. `100(sqrt(3+1))` m

Answer

To solve the problem step by step, we will use the concepts of trigonometry, specifically the tangent of angles in right triangles.

### Step 1: Understanding the problem
We have a tower (AB) and a person (C) standing 100 meters away from its base (B). The angle of elevation to the top of the tower (A) from point C is 45 degrees. We need to find out how much higher the tower must be raised so that the angle of elevation from the same point (C) becomes 60 degrees.

### Step 2: Setting up the first triangle
Let the height of the unfinished tower be \( H \). From triangle ABC, we can use the tangent of the angle of elevation:
\[ \tan(45^\circ) = \frac{H}{BC} \]
Here, \( BC = 100 \) meters and \( \tan(45^\circ) = 1 \). Therefore, we have:
\[ 1 = \frac{H}{100} \]
This implies:
\[ H = 100 \text{ meters} \]

### Step 3: Setting up the second triangle
Now, we consider the new height of the tower after it has been raised. Let the additional height be \( X \). The new height of the tower will be \( H + X \). From triangle DBC, where D is the new top of the tower, we can use the tangent for the angle of elevation of 60 degrees:
\[ \tan(60^\circ) = \frac{H + X}{BC} \]
Again, \( BC = 100 \) meters and \( \tan(60^\circ) = \sqrt{3} \). Therefore, we have:
\[ \sqrt{3} = \frac{H + X}{100} \]
This implies:
\[ H + X = 100\sqrt{3} \]

### Step 4: Substitute the value of H
We already found that \( H = 100 \) meters. Substituting this into the equation gives:
\[ 100 + X = 100\sqrt{3} \]

### Step 5: Solve for X
Now, we can solve for \( X \):
\[ X = 100\sqrt{3} - 100 \]
Factoring out 100, we get:
\[ X = 100(\sqrt{3} - 1) \]

### Conclusion
Thus, the tower must be raised by \( 100(\sqrt{3} - 1) \) meters for the angle of elevation to become 60 degrees.