If the points A (7, 9), B (3, - 7) and C (-3, 3) are the vertices of a triangle then find the measure of `angle C`

Options

A. `30^(@)`
B. `45^(@)`
C. `60^(@)`
D. `90^(@)`

Answer

To find the measure of angle C in triangle ABC with vertices A(7, 9), B(3, -7), and C(-3, 3), we will use the distance formula to calculate the lengths of the sides of the triangle and then apply the cosine rule to find the angle.

### Step-by-Step Solution:

1. **Calculate the lengths of the sides of the triangle:**
- **Length AB:**
\[ AB = \sqrt{(x_B - x_A)^2 + (y_B - y_A)^2} \]
\[ AB = \sqrt{(3 - 7)^2 + (-7 - 9)^2} \]
\[ AB = \sqrt{(-4)^2 + (-16)^2} = \sqrt{16 + 256} = \sqrt{272} \]

- **Length BC:**
\[ BC = \sqrt{(x_C - x_B)^2 + (y_C - y_B)^2} \]
\[ BC = \sqrt{(-3 - 3)^2 + (3 - (-7))^2} \]
\[ BC = \sqrt{(-6)^2 + (10)^2} = \sqrt{36 + 100} = \sqrt{136} \]

- **Length AC:**
\[ AC = \sqrt{(x_C - x_A)^2 + (y_C - y_A)^2} \]
\[ AC = \sqrt{(-3 - 7)^2 + (3 - 9)^2} \]
\[ AC = \sqrt{(-10)^2 + (-6)^2} = \sqrt{100 + 36} = \sqrt{136} \]

2. **Use the cosine rule to find angle C:**
The cosine rule states:
\[ c^2 = a^2 + b^2 - 2ab \cdot \cos(C) \]
Here, let:
- \( a = BC = \sqrt{136} \)
- \( b = AC = \sqrt{136} \)
- \( c = AB = \sqrt{272} \)

Plugging in the values:
\[ (\sqrt{272})^2 = (\sqrt{136})^2 + (\sqrt{136})^2 - 2 \cdot \sqrt{136} \cdot \sqrt{136} \cdot \cos(C) \]
\[ 272 = 136 + 136 - 2 \cdot 136 \cdot \cos(C) \]
\[ 272 = 272 - 272 \cdot \cos(C) \]
\[ 0 = -272 \cdot \cos(C) \]
\[ \cos(C) = 0 \]

3. **Determine the angle:**
Since \( \cos(C) = 0 \), this means:
\[ C = 90^\circ \]

### Conclusion:
The measure of angle C is \( 90^\circ \).