In a `Delta ABC, if b=20, c=21and sin A =3/5,` then the value of a is

Options

A. 12
B. 13
C. 14
D. 15

Answer

To find the value of \( a \) in triangle \( ABC \) given \( b = 20 \), \( c = 21 \), and \( \sin A = \frac{3}{5} \), we can follow these steps:

### Step 1: Use the sine definition to find the sides
Given that \( \sin A = \frac{3}{5} \), we can interpret this in terms of a right triangle where:
- The opposite side (perpendicular) to angle \( A \) is \( 3k \)
- The hypotenuse is \( 5k \)

### Step 2: Find the base using Pythagorean theorem
Using the Pythagorean theorem:
\[ \text{Hypotenuse}^2 = \text{Perpendicular}^2 + \text{Base}^2 \]
Substituting the known values:
\[ (5k)^2 = (3k)^2 + \text{Base}^2 \]
This simplifies to:
\[ 25k^2 = 9k^2 + \text{Base}^2 \]
\[ \text{Base}^2 = 25k^2 - 9k^2 = 16k^2 \]
Taking the square root gives:
\[ \text{Base} = 4k \]

### Step 3: Find \( \cos A \)
Now, we can find \( \cos A \):
\[ \cos A = \frac{\text{Base}}{\text{Hypotenuse}} = \frac{4k}{5k} = \frac{4}{5} \]

### Step 4: Use the cosine rule
According to the cosine rule:
\[ \cos A = \frac{b^2 + c^2 - a^2}{2bc} \]
Substituting the known values:
\[ \frac{4}{5} = \frac{20^2 + 21^2 - a^2}{2 \cdot 20 \cdot 21} \]
Calculating \( 20^2 \) and \( 21^2 \):
\[ 20^2 = 400, \quad 21^2 = 441 \]
Thus:
\[ \frac{4}{5} = \frac{400 + 441 - a^2}{840} \]
This simplifies to:
\[ \frac{4}{5} \cdot 840 = 400 + 441 - a^2 \]
Calculating \( \frac{4 \cdot 840}{5} = 672 \):
\[ 672 = 841 - a^2 \]

### Step 5: Solve for \( a^2 \)
Rearranging gives:
\[ a^2 = 841 - 672 = 169 \]

### Step 6: Find \( a \)
Taking the square root:
\[ a = \sqrt{169} = 13 \]

Thus, the value of \( a \) is \( 13 \).