The angle of elevation of the top of an incomplete tower, at a point 40m away from its foot, is `45^(@)`. How much more high the tower must be constructed so that the angle of elevation of its top at the same point be `60^(@)`?

Answer

To solve the problem step by step, we will use trigonometric ratios, specifically the tangent function, which relates the angle of elevation to the height of the tower and the distance from the tower.

### Step 1: Understand the problem
We have a point D that is 40 meters away from the foot of the tower (point B). The angle of elevation to the top of the incomplete tower (point C) from point D is 45 degrees. We need to find out how much more height (x) is required to raise the tower so that the angle of elevation from point D to the top of the completed tower (point A) is 60 degrees.

### Step 2: Set up the first triangle (BCD)
In triangle BCD, where:
- BC is the height of the incomplete tower (let's denote it as h).
- BD is the distance from the foot of the tower to point D, which is 40 m.
- The angle of elevation at point D to point C is 45 degrees.

Using the tangent function:
\[ \tan(45^\circ) = \frac{BC}{BD} \]
Since \(\tan(45^\circ) = 1\), we have:
\[ 1 = \frac{h}{40} \]
Thus, we can find the height of the incomplete tower:
\[ h = 40 \text{ m} \]

### Step 3: Set up the second triangle (ACD)
Now, we consider the completed tower (AC) where:
- AC is the height of the completed tower, which is \(h + x\) (where x is the additional height we need to find).
- The angle of elevation at point D to point A is 60 degrees.

Using the tangent function again:
\[ \tan(60^\circ) = \frac{AC}{BD} \]
Since \(\tan(60^\circ) = \sqrt{3}\), we have:
\[ \sqrt{3} = \frac{h + x}{40} \]

### Step 4: Substitute the height of the incomplete tower
Now substitute \(h = 40\) m into the equation:
\[ \sqrt{3} = \frac{40 + x}{40} \]

### Step 5: Solve for x
Cross-multiplying gives:
\[ 40\sqrt{3} = 40 + x \]
Rearranging this gives:
\[ x = 40\sqrt{3} - 40 \]

### Step 6: Calculate the value of x
Now, we can calculate the numerical value of \(x\):
\[ x = 40(\sqrt{3} - 1) \]
Using the approximate value of \(\sqrt{3} \approx 1.732\):
\[ x \approx 40(1.732 - 1) = 40(0.732) \approx 29.28 \text{ m} \]

### Conclusion
Thus, the additional height that must be constructed to achieve an angle of elevation of 60 degrees is approximately **29.28 meters**.

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