In a `triangleABC"if c"=(a+b) sin theta and cos theta=(ksqrtab)/(a+b)," then "k=`

Options

A. `2 cos""C/2`
B. `2cos""B/2`
C. `2cos""A/2`
D. `cos""C/2`

Answer

To find the value of \( k \) in the given triangle \( ABC \) where \( c = (a + b) \sin \theta \) and \( \cos \theta = \frac{k \sqrt{ab}}{a + b} \), we can follow these steps:

### Step-by-Step Solution:

1. **Start with the given equations**:
\[ c = (a + b) \sin \theta \]
\[ \cos \theta = \frac{k \sqrt{ab}}{a + b} \]

2. **Use the identity for cosine**:
Recall that \( \cos^2 \theta + \sin^2 \theta = 1 \). We can express \( \cos \theta \) in terms of \( \sin \theta \):
\[ \cos \theta = \sqrt{1 - \sin^2 \theta} \]

3. **Substitute \( \sin \theta \)**:
From the first equation, we have:
\[ \sin \theta = \frac{c}{a + b} \]
Substitute this into the cosine identity:
\[ \cos \theta = \sqrt{1 - \left(\frac{c}{a + b}\right)^2} \]

4. **Set the two expressions for \( \cos \theta \) equal**:
\[ \sqrt{1 - \left(\frac{c}{a + b}\right)^2} = \frac{k \sqrt{ab}}{a + b} \]

5. **Square both sides**:
\[ 1 - \left(\frac{c}{a + b}\right)^2 = \left(\frac{k \sqrt{ab}}{a + b}\right)^2 \]

6. **Multiply through by \( (a + b)^2 \)** to eliminate the denominator:
\[ (a + b)^2 - c^2 = k^2 ab \]

7. **Rearranging gives**:
\[ (a + b)^2 - c^2 = k^2 ab \]

8. **Use the cosine rule**:
From the cosine rule, we know:
\[ c^2 = a^2 + b^2 - 2ab \cos C \]
Therefore:
\[ (a + b)^2 - (a^2 + b^2 - 2ab \cos C) = k^2 ab \]
Simplifying gives:
\[ 2ab + 2ab \cos C = k^2 ab \]

9. **Factor out \( ab \)**:
\[ 2(1 + \cos C) = k^2 \]

10. **Take the square root**:
\[ k = \sqrt{2(1 + \cos C)} \]

11. **Using the double angle formula**:
Recall that \( \cos 2\theta = 2\cos^2 \theta - 1 \). Thus:
\[ 1 + \cos C = 2\cos^2\left(\frac{C}{2}\right) \]
Therefore:
\[ k = 2\cos\left(\frac{C}{2}\right) \]

### Final Result:
Thus, the value of \( k \) is:
\[ k = 2 \cos\left(\frac{C}{2}\right) \]