In a `DeltaABC` if `/_A=/_B=1/2(sin^(-1)((sqrt(6)+1)/(2sqrt(3)))+sin^(-1)(1/(sqrt(3))))` and length of `c=6.3^(1/4)`, then the area of `DeltaABC`

Answer

To find the area of triangle \( \Delta ABC \) given the angles and the length of side \( c \), we can follow these steps:

### Step 1: Calculate the angles \( A \) and \( B \)
We are given:
\[ \angle A = \angle B = \frac{1}{2} \left( \sin^{-1} \left( \frac{\sqrt{6} + 1}{2\sqrt{3}} \right) + \sin^{-1} \left( \frac{1}{\sqrt{3}} \right) \right) \]

### Step 2: Use the sine addition formula
Using the formula for \( \sin^{-1}(x) + \sin^{-1}(y) \):
\[ \sin^{-1}(x) + \sin^{-1}(y) = \sin^{-1} \left( x\sqrt{1-y^2} + y\sqrt{1-x^2} \right) \]
Let \( x = \frac{\sqrt{6} + 1}{2\sqrt{3}} \) and \( y = \frac{1}{\sqrt{3}} \).

### Step 3: Calculate \( \sqrt{1 - y^2} \) and \( \sqrt{1 - x^2} \)
Calculate \( \sqrt{1 - y^2} \):
\[ y^2 = \left( \frac{1}{\sqrt{3}} \right)^2 = \frac{1}{3} \implies \sqrt{1 - y^2} = \sqrt{1 - \frac{1}{3}} = \sqrt{\frac{2}{3}} = \frac{\sqrt{6}}{3} \]

Calculate \( \sqrt{1 - x^2} \):
\[ x^2 = \left( \frac{\sqrt{6} + 1}{2\sqrt{3}} \right)^2 = \frac{7 + 2\sqrt{6}}{12} \implies 1 - x^2 = 1 - \frac{7 + 2\sqrt{6}}{12} = \frac{5 - 2\sqrt{6}}{12} \]
Thus, \( \sqrt{1 - x^2} = \sqrt{\frac{5 - 2\sqrt{6}}{12}} \).

### Step 4: Substitute values into the sine addition formula
Now substitute back into the sine addition formula:
\[ \sin^{-1} \left( \frac{\sqrt{6} + 1}{2\sqrt{3}} \cdot \frac{\sqrt{6}}{3} + \frac{1}{\sqrt{3}} \cdot \sqrt{\frac{5 - 2\sqrt{6}}{12}} \right) \]

### Step 5: Calculate \( \angle C \)
Since \( \angle A + \angle B + \angle C = 180^\circ \) and \( \angle A = \angle B \), we can find \( \angle C \).

### Step 6: Calculate the area of triangle \( ABC \)
For an isosceles triangle:
\[ \text{Area} = \frac{1}{2}ab \sin C \]
where \( a = b = c \) and \( c = 6.3^{1/4} \).

### Step 7: Substitute the values
Using the area formula for an equilateral triangle:
\[ \text{Area} = \frac{\sqrt{3}}{4} s^2 \]
where \( s = 6.3^{1/4} \).

### Step 8: Final area calculation
\[ \text{Area} = \frac{\sqrt{3}}{4} \left( 6 \cdot 3^{1/4} \right)^2 = \frac{\sqrt{3}}{4} \cdot 36 \cdot 3^{1/2} = \frac{\sqrt{3}}{4} \cdot 36 \cdot \sqrt{3} = \frac{108}{4} = 27 \]

Thus, the area of triangle \( \Delta ABC \) is:
\[ \boxed{27} \]