Find the tangents of the angles in triangle with vertices A(1, 1), B(5,2) and C(3, 5).


To find the tangents of the angles in a triangle with given vertices \( A(1, 1) \), \( B(5, 2) \), and \( C(3, 5) \), we will follow these steps:

### Step 1: Calculate the lengths of the sides of the triangle.
The length of a side between two points can be found using the distance formula:
\[
d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
\]

- **Side \( AB \)**:
\[
AB = \sqrt{(5 - 1)^2 + (2 - 1)^2} = \sqrt{16 + 1} = \sqrt{17}
\]
- **Side \( BC \)**:
\[
BC = \sqrt{(3 - 5)^2 + (5 - 2)^2} = \sqrt{4 + 9} = \sqrt{13}
\]
- **Side \( CA \)**:
\[
CA = \sqrt{(3 - 1)^2 + (5 - 1)^2} = \sqrt{4 + 16} = \sqrt{20}
\]

### Step 2: Use the tangent of the angle formula.
For a triangle, the tangent of an angle can be found using the formula derived from the Law of Cosines:
\[
\tan(\theta) = \frac{\text{opposite side}}{\text{adjacent sides}}
\]
However, since we need the tangent of each angle, we will use the following identity:
\[
\tan(\angle A) = \frac{\text{Area of the triangle} \times 4}{AB \times CA}
\]
We can first calculate the area of the triangle and then use it to find the tangents of the angles.

### Step 3: Calculate the area of the triangle.
The area of a triangle with vertices \( (x_1, y_1), (x_2, y_2), (x_3, y_3) \) is given by:
\[
\text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|
\]
Substituting the coordinates of points \( A(1, 1) \), \( B(5, 2) \), and \( C(3, 5) \):
\[
\text{Area} = \frac{1}{2} \left| 1(2 - 5) + 5(5 - 1) + 3(1 - 2) \right|
\]
\[
\text{Area} = \frac{1}{2} \left| 1(-3) + 5(4) + 3(-1) \right| = \frac{1}{2} \left| -3 + 20 - 3 \right| = \frac{1}{2} \times 14 = 7
\]

### Step 4: Find the tangents of the angles.
Now we use the formula:
\[
\tan(\theta_A) = \frac{4 \times \text{Area}}{AB \times CA}
\]
- **For angle \( A \)**:
\[
\tan(\theta_A) = \frac{4 \times 7}{\sqrt{17} \times \sqrt{20}} = \frac{28}{\sqrt{340}} = \frac{28}{18.44} \approx 1.52
\]
- **For angle \( B \)**:
\[
\tan(\theta_B) = \frac{4 \times 7}{\sqrt{13} \times \sqrt{17}} = \frac{28}{\sqrt{221}} = \frac{28}{14.87} \approx 1.88
\]
- **For angle \( C \)**:
\[
\tan(\theta_C) = \frac{4 \times 7}{\sqrt{20} \times \sqrt{13}} = \frac{28}{\sqrt{260}} = \frac{28}{16.12} \approx 1.74
\]

Thus, the tangents of the angles are approximately:
- \( \tan(\theta_A) \approx 1.52 \)
- \( \tan(\theta_B) \approx 1.88 \)
- \( \tan(\theta_C) \approx 1.74 \)

Would you like to discuss these calculations in more detail or explore another concept?

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**Related Questions:**
1. How can the Law of Cosines be used to find angles in a triangle?
2. What is the significance of the tangent of an angle in a triangle?
3. How can the sine rule help in solving triangles?
4. What is Heron's formula for the area of a triangle, and how does it compare with the determinant method used here?
5. How do the properties of a right triangle affect the values of trigonometric functions?

**Tip:** Remember that the sum of angles in any triangle is always 180°, which can help you verify your results!

Find the tangents of the angles in triangle with vertices A(1, 1), B(5,2) and C(3, 5).

Discover how to calculate the tangents of angles in a triangle with vertices A(1, 1), B(5, 2), and C(3, 5). This solution covers distance formula, the determinant method for the area, and tangent calculations for each angle. Suitable for students in Grades 10-12.

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