In an acute-angled triangle ABC if sin `(B + C - A) = (sqrt3)/2` and `tan (C+A-B) = 1`, then C is equal to

Options

A. `37.5^@`
B. `67.5^@`
C. `52.5^@`
D. `72.5^@`

Answer

To solve the problem, we need to analyze the given equations step by step.

### Step 1: Analyze the given equations
We have two equations based on the given conditions:
1. \( \sin(B + C - A) = \frac{\sqrt{3}}{2} \)
2. \( \tan(C + A - B) = 1 \)

### Step 2: Relate the sine and tangent values to angles
From trigonometric values, we know:
- \( \sin(60^\circ) = \frac{\sqrt{3}}{2} \)
- \( \tan(45^\circ) = 1 \)

Thus, we can set up the equations:
1. \( B + C - A = 60^\circ \) (Equation 1)
2. \( C + A - B = 45^\circ \) (Equation 2)

### Step 3: Rearranging the equations
From Equation 1:
\[ B + C - A = 60^\circ \]
Rearranging gives:
\[ -A + B + C = 60^\circ \]

From Equation 2:
\[ C + A - B = 45^\circ \]
Rearranging gives:
\[ A - B + C = 45^\circ \]

### Step 4: Adding the two equations
Now, we can add the two rearranged equations:
\[ (-A + B + C) + (A - B + C) = 60^\circ + 45^\circ \]
This simplifies to:
\[ 2C = 105^\circ \]

### Step 5: Solve for C
Dividing both sides by 2 gives:
\[ C = \frac{105^\circ}{2} = 52.5^\circ \]

### Conclusion
Thus, the value of \( C \) is \( 52.5^\circ \).

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