In a `DeltaABC, 2 cos A=(sin B)/(sin C) and 2 ^(tan^(2)B)` is a solution of equation `x^(2)-9x +8=0,` then `DeltaABC` is

Options

A. equilateral
B. isosceles
C. scalene
D. right angled

Answer

To solve the problem step by step, we will analyze the given conditions and apply the properties of triangles.

### Step 1: Analyze the given equation
We start with the equation provided in the problem:
\[ 2 \cos A = \frac{\sin B}{\sin C} \]

### Step 2: Use the Sine Rule
According to the Sine Rule:
\[ \frac{A}{\sin A} = \frac{B}{\sin B} = \frac{C}{\sin C} \]
From this, we can express \(\frac{\sin B}{\sin C}\) as:
\[ \frac{\sin B}{\sin C} = \frac{B}{C} \]

### Step 3: Substitute into the equation
Substituting this into our original equation gives:
\[ 2 \cos A = \frac{B}{C} \]

### Step 4: Use the Cosine Rule
We know from the Cosine Rule:
\[ \cos A = \frac{B^2 + C^2 - A^2}{2BC} \]
Thus, substituting this into our equation:
\[ 2 \left(\frac{B^2 + C^2 - A^2}{2BC}\right) = \frac{B}{C} \]
This simplifies to:
\[ \frac{B^2 + C^2 - A^2}{BC} = \frac{B}{C} \]

### Step 5: Cross-multiply
Cross-multiplying gives:
\[ B^2 + C^2 - A^2 = B \cdot \frac{BC}{C} \]
This simplifies to:
\[ B^2 + C^2 - A^2 = B^2 \]

### Step 6: Rearranging the equation
Rearranging the equation leads to:
\[ C^2 - A^2 = 0 \]
This implies:
\[ C^2 = A^2 \]
Thus:
\[ C = A \]

### Step 7: Determine the implications
Since \(C = A\), this indicates that triangle ABC is isosceles with \(A = C\).

### Step 8: Analyze the second condition
Next, we analyze the second condition:
\[ 2^{\tan^2 B} \text{ is a solution of } x^2 - 9x + 8 = 0 \]

### Step 9: Solve the quadratic equation
The quadratic can be factored as:
\[ (x - 1)(x - 8) = 0 \]
Thus, the solutions are:
\[ x = 1 \quad \text{or} \quad x = 8 \]

### Step 10: Set the solutions equal to \(2^{\tan^2 B}\)
This gives us two cases:
1. \(2^{\tan^2 B} = 1\) which implies \(\tan^2 B = 0\) (not possible for a triangle).
2. \(2^{\tan^2 B} = 8\) which implies \(\tan^2 B = 3\).

### Step 11: Find angle B
Taking the square root gives:
\[ \tan B = \sqrt{3} \]
This means:
\[ B = 60^\circ \text{ (since angles in a triangle are acute)} \]

### Step 12: Find angles A and C
Since \(C = A\) and \(A + B + C = 180^\circ\):
Let \(A = C\), then:
\[ 2A + 60^\circ = 180^\circ \]
This simplifies to:
\[ 2A = 120^\circ \]
Thus:
\[ A = 60^\circ \quad \text{and} \quad C = 60^\circ \]

### Conclusion
Therefore, triangle ABC is an equilateral triangle with all angles equal to \(60^\circ\).