If, in a `Delta ABC, a = 6cm , b = 8 cm, c = 10cm,`
then : `sin 2A` =

Options

A. `6/25`
B. `8/25`
C. `10/25`
D. `24/25`

Answer

To find the value of \( \sin 2A \) in triangle \( ABC \) where \( a = 6 \, \text{cm} \), \( b = 8 \, \text{cm} \), and \( c = 10 \, \text{cm} \), we can follow these steps:

### Step 1: Use the formula for \( \sin 2A \)
We know that:
\[ \sin 2A = 2 \sin A \cos A \]
So, we need to find \( \sin A \) and \( \cos A \).

### Step 2: Use the Cosine Rule to find \( \cos A \)
The Cosine Rule states:
\[ \cos A = \frac{b^2 + c^2 - a^2}{2bc} \]
Substituting the values:
\[ \cos A = \frac{8^2 + 10^2 - 6^2}{2 \cdot 8 \cdot 10} \]
Calculating the squares:
\[ \cos A = \frac{64 + 100 - 36}{160} = \frac{128}{160} \]
Now simplify:
\[ \cos A = \frac{64}{80} = \frac{32}{40} = \frac{4}{5} \]

### Step 3: Use the Pythagorean identity to find \( \sin A \)
We know:
\[ \sin^2 A + \cos^2 A = 1 \]
Substituting \( \cos A \):
\[ \sin^2 A + \left(\frac{4}{5}\right)^2 = 1 \]
Calculating \( \left(\frac{4}{5}\right)^2 \):
\[ \sin^2 A + \frac{16}{25} = 1 \]
Subtracting \( \frac{16}{25} \) from both sides:
\[ \sin^2 A = 1 - \frac{16}{25} = \frac{25}{25} - \frac{16}{25} = \frac{9}{25} \]
Taking the square root:
\[ \sin A = \frac{3}{5} \]

### Step 4: Substitute \( \sin A \) and \( \cos A \) into the formula for \( \sin 2A \)
Now we can find \( \sin 2A \):
\[ \sin 2A = 2 \sin A \cos A = 2 \cdot \frac{3}{5} \cdot \frac{4}{5} \]
Calculating this:
\[ \sin 2A = 2 \cdot \frac{12}{25} = \frac{24}{25} \]

### Final Answer
Thus, the value of \( \sin 2A \) is:
\[ \sin 2A = \frac{24}{25} \]