In `triangle ABC, angleA=pi/3 and b:c =2:3, tan theta=sqrt3/5, 0 lt theta lt pi/2` then

Options

A. `B=60^(@)+0`
B. `C=60^(@)+theta`
C. `C=60^(@)-theta`
D. `C=60^(@)-theta`

Answer

To solve the problem step by step, we will use the given information about triangle ABC and apply trigonometric identities and properties of triangles.

### Step 1: Understand the given information
We are given:
- Angle A = π/3 (which is 60 degrees)
- The ratio of sides b:c = 2:3
- tan(θ) = √3/5, where 0 < θ < π/2

### Step 2: Use the angle sum property of triangles
In triangle ABC, the sum of angles is 180 degrees. Therefore, we have:
\[ A + B + C = 180^\circ \]
Substituting the value of A:
\[ \frac{\pi}{3} + B + C = 180^\circ \]

### Step 3: Express B + C in terms of A
Rearranging gives:
\[ B + C = 180^\circ - \frac{\pi}{3} \]
Converting π/3 to degrees:
\[ 180^\circ - 60^\circ = 120^\circ \]
Thus:
\[ B + C = 120^\circ \]

### Step 4: Use the given ratio b:c
Let b = 2k and c = 3k for some k. According to the Law of Sines:
\[ \frac{b}{\sin B} = \frac{c}{\sin C} \]
Substituting the values:
\[ \frac{2k}{\sin B} = \frac{3k}{\sin C} \]
This simplifies to:
\[ \frac{2}{\sin B} = \frac{3}{\sin C} \]
Cross-multiplying gives:
\[ 2 \sin C = 3 \sin B \]

### Step 5: Use the tangent difference formula
We know from the problem statement that:
\[ \tan\left(\frac{C - B}{2}\right) = \frac{C - B}{C + B} \]
Substituting the values of C and B:
\[ \tan\left(\frac{C - B}{2}\right) = \frac{C - B}{120^\circ} \]

### Step 6: Substitute values to find C - B
Using the ratio b:c and the known angles, we can express C - B in terms of θ:
From the problem, we have:
\[ \tan\left(\frac{C - B}{2}\right) = \frac{1}{5\sqrt{3}} \]
Setting this equal to tan(θ):
\[ \frac{C - B}{120^\circ} = \frac{1}{5\sqrt{3}} \]
Cross-multiplying gives:
\[ C - B = \frac{120^\circ}{5\sqrt{3}} \]

### Step 7: Find C and B
Now we have two equations:
1. \( C + B = 120^\circ \)
2. \( C - B = \frac{120^\circ}{5\sqrt{3}} \)

Adding these two equations:
\[ 2C = 120^\circ + \frac{120^\circ}{5\sqrt{3}} \]
Thus:
\[ C = 60^\circ + \frac{60^\circ}{5\sqrt{3}} \]

### Step 8: Conclusion
Finally, we can express C in terms of θ:
\[ C = 60^\circ + θ \]
Thus, the answer is:
\[ C = 60 + θ \]