The tangents from `(1, 2sqrt2)` to the hyperbola `16x^2-25y^2 = 400` include between them an angle equal to:

Explanation

To solve the problem of finding the angle between the tangents from the point \( (1, 2\sqrt{2}) \) to the hyperbola given by the equation \( 16x^2 - 25y^2 = 400 \), we can follow these steps:

### Step 1: Rewrite the Hyperbola Equation
First, we rewrite the hyperbola equation in standard form. The given equation is:

\[ 16x^2 - 25y^2 = 400 \]

Dividing through by 400 gives:

\[ \frac{x^2}{25} - \frac{y^2}{16} = 1 \]

Here, we identify \( a^2 = 25 \) and \( b^2 = 16 \), which means \( a = 5 \) and \( b = 4 \).

### Step 2: Use the Tangent Equation
The equation of the tangent to the hyperbola at a point \( (x_0, y_0) \) is given by:

\[ \frac{xx_0}{a^2} - \frac{yy_0}{b^2} = 1 \]

For our point \( (1, 2\sqrt{2}) \), we substitute \( x_0 = 1 \) and \( y_0 = 2\sqrt{2} \):

\[ \frac{x \cdot 1}{25} - \frac{y \cdot 2\sqrt{2}}{16} = 1 \]

### Step 3: Rearranging the Tangent Equation
Rearranging gives us:

\[ \frac{x}{25} - \frac{2\sqrt{2}y}{16} = 1 \]

Multiplying through by 400 (the least common multiple of 25 and 16) to eliminate the fractions:

\[ 16x - 50\sqrt{2}y = 400 \]

### Step 4: Finding the Slopes of the Tangents
The slope \( m \) of the tangent lines can be found by rewriting the equation in slope-intercept form \( y = mx + c \):

\[ 50\sqrt{2}y = 16x - 400 \]
\[ y = \frac{16}{50\sqrt{2}}x - \frac{400}{50\sqrt{2}} \]

The slope \( m \) is:

\[ m = \frac{16}{50\sqrt{2}} = \frac{8}{25\sqrt{2}} \]

### Step 5: Finding the Angle Between the Tangents
For a hyperbola, if the slopes of the tangents are \( m_1 \) and \( m_2 \), the angle \( \theta \) between the tangents can be found using the formula:

\[ \tan \theta = \frac{m_1 - m_2}{1 + m_1 m_2} \]

Since the tangents are symmetrical with respect to the center of the hyperbola, we can conclude that \( m_1 \cdot m_2 = -1 \) (the product of the slopes of perpendicular lines).

### Step 6: Conclusion
Since \( m_1 m_2 = -1 \), we can substitute this into our angle formula:

\[ \tan \theta = \frac{m_1 - m_2}{1 - 1} \rightarrow \text{undefined} \]

This indicates that the angle \( \theta \) is \( \frac{\pi}{2} \) radians (or 90 degrees).

Thus, the angle between the tangents from the point \( (1, 2\sqrt{2}) \) to the hyperbola \( 16x^2 - 25y^2 = 400 \) is:

\[ \boxed{\frac{\pi}{2}} \]