Let `alpha, beta` are the root of equation `acostheta + bsintheta = c`.
If `alpha = 30^(@)` and `beta = 60^(@)` such that `a, b, c` represent sides of a `DeltaABC` then

Options

A. `ABC` is acute angle triangle
B. `ABC` is acute isosceles triangle
C. `ABC` is right angle triangle
D. `ABC` is obtuse angle triangle

Explanation

To solve the problem step by step, we will analyze the given equation and the roots provided.

### Step 1: Write the equation and substitute the roots
The given equation is:
\[ a \cos \theta + b \sin \theta = c \]
We know that the roots of this equation are \( \alpha = 30^\circ \) and \( \beta = 60^\circ \).

### Step 2: Substitute \( \alpha = 30^\circ \) into the equation
Substituting \( \alpha \) into the equation:
\[ a \cos(30^\circ) + b \sin(30^\circ) = c \]
Using the values of cosine and sine:
\[ a \left(\frac{\sqrt{3}}{2}\right) + b \left(\frac{1}{2}\right) = c \]
This simplifies to:
\[ \frac{\sqrt{3}}{2} a + \frac{1}{2} b = c \]
Let's call this Equation (1).

### Step 3: Substitute \( \beta = 60^\circ \) into the equation
Now substituting \( \beta \) into the equation:
\[ a \cos(60^\circ) + b \sin(60^\circ) = c \]
Using the values of cosine and sine:
\[ a \left(\frac{1}{2}\right) + b \left(\frac{\sqrt{3}}{2}\right) = c \]
This simplifies to:
\[ \frac{1}{2} a + \frac{\sqrt{3}}{2} b = c \]
Let's call this Equation (2).

### Step 4: Set Equations (1) and (2) equal to each other
Since both equations equal \( c \), we can set them equal to each other:
\[ \frac{\sqrt{3}}{2} a + \frac{1}{2} b = \frac{1}{2} a + \frac{\sqrt{3}}{2} b \]

### Step 5: Rearrange the equation
Rearranging gives:
\[ \frac{\sqrt{3}}{2} a - \frac{1}{2} a = \frac{\sqrt{3}}{2} b - \frac{1}{2} b \]
This simplifies to:
\[ \left(\frac{\sqrt{3}}{2} - \frac{1}{2}\right) a = \left(\frac{\sqrt{3}}{2} - \frac{1}{2}\right) b \]

### Step 6: Factor out common terms
Factoring out the common term:
\[ \left(\frac{\sqrt{3} - 1}{2}\right) a = \left(\frac{\sqrt{3} - 1}{2}\right) b \]

### Step 7: Solve for \( a \) and \( b \)
Since \( \frac{\sqrt{3} - 1}{2} \) is not zero, we can divide both sides by it:
\[ a = b \]
This indicates that triangle \( ABC \) is isosceles.

### Step 8: Determine the nature of triangle \( ABC \)
Now we need to check the angles of triangle \( ABC \). We can use the cosine rule:
\[ \cos C = \frac{a^2 + b^2 - c^2}{2ab} \]
Since \( a = b \), we can substitute:
\[ \cos C = \frac{2a^2 - c^2}{2a^2} \]

### Step 9: Substitute \( c \) in terms of \( a \)
From either Equation (1) or (2), substituting \( c \):
Using Equation (1):
\[ c = \frac{\sqrt{3}}{2} a + \frac{1}{2} b \]
Since \( a = b \):
\[ c = \frac{\sqrt{3}}{2} a + \frac{1}{2} a = \left(\frac{\sqrt{3} + 1}{2}\right) a \]

### Step 10: Substitute \( c \) back into the cosine formula
Now substituting \( c \) into the cosine formula:
\[ \cos C = \frac{2a^2 - \left(\frac{\sqrt{3} + 1}{2} a\right)^2}{2a^2} \]

### Step 11: Simplify the expression
Calculating the square:
\[ \left(\frac{\sqrt{3} + 1}{2}\right)^2 = \frac{(\sqrt{3} + 1)^2}{4} = \frac{3 + 2\sqrt{3} + 1}{4} = \frac{4 + 2\sqrt{3}}{4} = 1 + \frac{\sqrt{3}}{2} \]
Thus:
\[ \cos C = \frac{2a^2 - \left(1 + \frac{\sqrt{3}}{2}\right)a^2}{2a^2} = \frac{(2 - 1 - \frac{\sqrt{3}}{2})a^2}{2a^2} = \frac{1 - \frac{\sqrt{3}}{2}}{2} \]

### Step 12: Determine the angle type
Since \( \cos C > 0 \), angle \( C \) is acute.

### Conclusion
Thus, triangle \( ABC \) is an **acute isosceles triangle**.