If the sides of a triangle are 6cm, 10cm and 14cm, then what is the largest angle included by the sides?

Options

A. A. `90^(@)`
B. B. `120^(@)`
C. C. `135^(@)`
D. D. `150^(@)`

Answer

To find the largest angle included by the sides of a triangle with lengths 6 cm, 10 cm, and 14 cm, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the sides of the triangle**:
Let the sides be:
- \( A = 6 \, \text{cm} \)
- \( B = 10 \, \text{cm} \)
- \( C = 14 \, \text{cm} \)

2. **Determine the largest side**:
The largest angle in a triangle is opposite the longest side. Here, the longest side is \( C = 14 \, \text{cm} \).

3. **Use the Law of Cosines**:
To find the angle opposite the longest side (angle \( C \)), we can use the Law of Cosines:
\[ C^2 = A^2 + B^2 - 2AB \cdot \cos(C) \]
Rearranging the formula to find \( \cos(C) \):
\[ \cos(C) = \frac{A^2 + B^2 - C^2}{2AB} \]

4. **Substitute the values into the formula**:
\[ \cos(C) = \frac{6^2 + 10^2 - 14^2}{2 \cdot 6 \cdot 10} \]
Calculate \( 6^2 \), \( 10^2 \), and \( 14^2 \):
\[ 6^2 = 36, \quad 10^2 = 100, \quad 14^2 = 196 \]
Now substitute these values:
\[ \cos(C) = \frac{36 + 100 - 196}{120} = \frac{-60}{120} = -0.5 \]

5. **Find angle \( C \)**:
Now, we need to find the angle whose cosine is \(-0.5\). This corresponds to:
\[ C = \cos^{-1}(-0.5) \]
The angle \( C \) can be \( 120^\circ \) (since cosine is negative in the second quadrant).

6. **Conclusion**:
The largest angle included by the sides of the triangle is:
\[ \boxed{120^\circ} \]