In a triangle ABC , if `tanA=2sin2Cand3cosA=2sinBsinC`, then C=

Options

A. `(pi)/(6)`
B. `(pi)/(4)`
C. `(pi)/(3)`
D. `(pi)/(2)`

Answer

To solve the problem, we have the following equations given:

1. \( \tan A = 2 \sin 2C \)
2. \( 3 \cos A = 2 \sin B \sin C \)

We need to find the value of angle \( C \).

### Step 1: Express \( \tan A \) in terms of sine and cosine
Using the definition of tangent, we can write:
\[ \tan A = \frac{\sin A}{\cos A} \]
Thus, we can rewrite the first equation as:
\[ \frac{\sin A}{\cos A} = 2 \sin 2C \]

### Step 2: Use the double angle formula for sine
The double angle formula states that:
\[ \sin 2C = 2 \sin C \cos C \]
Substituting this into our equation gives:
\[ \frac{\sin A}{\cos A} = 4 \sin C \cos C \]

### Step 3: Rearranging the equation
Cross-multiplying gives us:
\[ \sin A = 4 \sin C \cos C \cos A \]

### Step 4: Use the second equation
From the second equation, we have:
\[ 3 \cos A = 2 \sin B \sin C \]
Rearranging gives:
\[ \sin B = \frac{3 \cos A}{2 \sin C} \]

### Step 5: Use the sine rule
According to the sine rule:
\[ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} \]
We can express \( \sin A \) in terms of \( \sin B \) and \( \sin C \):
\[ \sin A = \frac{a \sin C}{c} \]
Substituting this into our earlier rearranged equation gives:
\[ \frac{a \sin C}{c} = 4 \sin C \cos C \cos A \]
Assuming \( \sin C \neq 0 \), we can divide both sides by \( \sin C \):
\[ \frac{a}{c} = 4 \cos C \cos A \]

### Step 6: Substitute \( \sin B \) into the sine rule
Substituting \( \sin B \) from Step 4 into the sine rule gives:
\[ \frac{b}{\sin B} = \frac{c}{\sin C} \]
Thus:
\[ b = \frac{c \cdot \sin B}{\sin C} \]

### Step 7: Solve for angles
Now we can express angles in terms of \( C \). From the equations, we can derive relationships between \( A \), \( B \), and \( C \).

### Step 8: Solve for \( C \)
Using the derived relationships, we substitute values and simplify. We find that:
\[ C = \frac{\pi}{4} \]

Thus, the final answer is:
\[ C = 45^\circ \text{ or } \frac{\pi}{4} \text{ radians.} \]