In a `Delta ABC` with vertices A(1,2), B(2,3) and C(3, 1) and `angle A = angle B = cos^(-1)((1)/(sqrt(10))), angle C = cos^(-1)((4)/(5))`, then find the circumentre of `Delta ABC`.

Answer

To find the circumcenter of triangle ABC with vertices A(1, 2), B(2, 3), and C(3, 1), we will use the formula for the circumcenter based on the angles of the triangle. The circumcenter (O) can be calculated using the following formulas:

\[ O_x = \frac{x_1 \sin 2A + x_2 \sin 2B + x_3 \sin 2C}{\sin 2A + \sin 2B + \sin 2C} \]
\[ O_y = \frac{y_1 \sin 2A + y_2 \sin 2B + y_3 \sin 2C}{\sin 2A + \sin 2B + \sin 2C} \]

### Step 1: Calculate the sine values of angles A, B, and C

Given:
- \( \cos A = \frac{1}{\sqrt{10}} \)
- \( \cos B = \frac{1}{\sqrt{10}} \)
- \( \cos C = \frac{4}{5} \)

Using the Pythagorean identity \( \sin^2 \theta + \cos^2 \theta = 1 \):

1. For angle A:
\[ \sin A = \sqrt{1 - \left(\frac{1}{\sqrt{10}}\right)^2} = \sqrt{1 - \frac{1}{10}} = \sqrt{\frac{9}{10}} = \frac{3}{\sqrt{10}} \]

2. For angle B (same as angle A):
\[ \sin B = \frac{3}{\sqrt{10}} \]

3. For angle C:
\[ \sin C = \sqrt{1 - \left(\frac{4}{5}\right)^2} = \sqrt{1 - \frac{16}{25}} = \sqrt{\frac{9}{25}} = \frac{3}{5} \]

### Step 2: Calculate sine of double angles

Using the double angle formula \( \sin 2\theta = 2 \sin \theta \cos \theta \):

1. For angle A:
\[ \sin 2A = 2 \sin A \cos A = 2 \cdot \frac{3}{\sqrt{10}} \cdot \frac{1}{\sqrt{10}} = \frac{6}{10} = \frac{3}{5} \]

2. For angle B (same as angle A):
\[ \sin 2B = \frac{3}{5} \]

3. For angle C:
\[ \sin 2C = 2 \sin C \cos C = 2 \cdot \frac{3}{5} \cdot \frac{4}{5} = \frac{24}{25} \]

### Step 3: Substitute values into the circumcenter formulas

Coordinates of points:
- \( A(1, 2) \)
- \( B(2, 3) \)
- \( C(3, 1) \)

Substituting into the formulas:

#### For \( O_x \):
\[ O_x = \frac{1 \cdot \frac{3}{5} + 2 \cdot \frac{3}{5} + 3 \cdot \frac{24}{25}}{\frac{3}{5} + \frac{3}{5} + \frac{24}{25}} \]

Calculating the numerator:
\[ = \frac{3}{5} + \frac{6}{5} + \frac{72}{25} = \frac{15}{25} + \frac{30}{25} + \frac{72}{25} = \frac{117}{25} \]

Calculating the denominator:
\[ = \frac{3}{5} + \frac{3}{5} + \frac{24}{25} = \frac{15}{25} + \frac{15}{25} + \frac{24}{25} = \frac{54}{25} \]

Thus,
\[ O_x = \frac{\frac{117}{25}}{\frac{54}{25}} = \frac{117}{54} = \frac{39}{18} = \frac{13}{6} \]

#### For \( O_y \):
\[ O_y = \frac{2 \cdot \frac{3}{5} + 3 \cdot \frac{3}{5} + 1 \cdot \frac{24}{25}}{\frac{3}{5} + \frac{3}{5} + \frac{24}{25}} \]

Calculating the numerator:
\[ = \frac{6}{5} + \frac{9}{5} + \frac{24}{25} = \frac{30}{25} + \frac{45}{25} + \frac{24}{25} = \frac{99}{25} \]

Thus,
\[ O_y = \frac{\frac{99}{25}}{\frac{54}{25}} = \frac{99}{54} = \frac{33}{18} = \frac{11}{6} \]

### Final Result

The circumcenter of triangle ABC is:
\[ O\left(\frac{13}{6}, \frac{11}{6}\right) \]