In a `DeltaABC,` if `a=13, b=14and c=15, ` then `angle A` is equal to (All symbols used have their usual meaning in a triangle.)

Options

A. `sin ^(-1)""4/5`
B. `sin ^(-1)""3/5`
C. `sin ^(-1)""3/4`
D. `sin ^(-1)""2/3`

Answer

To find angle A in triangle ABC with sides \( a = 13 \), \( b = 14 \), and \( c = 15 \), we can use the cosine rule. The cosine rule states:

\[ \cos A = \frac{b^2 + c^2 - a^2}{2bc} \]

### Step-by-step Solution:

1. **Identify the sides**:
- \( a = 13 \) (side opposite angle A)
- \( b = 14 \) (side opposite angle B)
- \( c = 15 \) (side opposite angle C)

2. **Substitute the values into the cosine rule**:
\[ \cos A = \frac{b^2 + c^2 - a^2}{2bc} \]
Substituting the values:
\[ \cos A = \frac{14^2 + 15^2 - 13^2}{2 \cdot 14 \cdot 15} \]

3. **Calculate \( b^2 \), \( c^2 \), and \( a^2 \)**:
- \( 14^2 = 196 \)
- \( 15^2 = 225 \)
- \( 13^2 = 169 \)

4. **Plug these values back into the equation**:
\[ \cos A = \frac{196 + 225 - 169}{2 \cdot 14 \cdot 15} \]

5. **Simplify the numerator**:
\[ 196 + 225 - 169 = 252 \]
So,
\[ \cos A = \frac{252}{2 \cdot 14 \cdot 15} \]

6. **Calculate \( 2bc \)**:
\[ 2 \cdot 14 \cdot 15 = 420 \]

7. **Now substitute this back into the equation**:
\[ \cos A = \frac{252}{420} \]

8. **Simplify \( \frac{252}{420} \)**:
- Dividing both the numerator and the denominator by 84 gives:
\[ \cos A = \frac{3}{5} \]

9. **Find \( \sin A \)** using the identity \( \sin^2 A + \cos^2 A = 1 \):
\[ \sin^2 A = 1 - \cos^2 A = 1 - \left(\frac{3}{5}\right)^2 = 1 - \frac{9}{25} = \frac{16}{25} \]

10. **Take the square root**:
\[ \sin A = \frac{4}{5} \]

11. **Find angle A**:
\[ A = \sin^{-1}\left(\frac{4}{5}\right) \]

### Final Answer:
\[ \text{Angle A} = \sin^{-1}\left(\frac{4}{5}\right) \]