Circumradius of an isosceles `DeltaABC` with `angleA=angleB` is 4 times its in radius, then cosA is root of the equation :

Options

A. `x^(2)-x-8=0`
B. `8x^(2)-8x+1=0`
C. `x^(2)-x-4=0`
D. `4x^(2)-4x+1=0`

Answer

To solve the problem, we need to find the equation whose root is \( \cos A \) given that the circumradius \( R \) of an isosceles triangle \( \Delta ABC \) with \( \angle A = \angle B \) is 4 times its inradius \( r \).

### Step-by-step Solution:

1. **Write the given relationship**:
\[ R = 4r \]

2. **Use the formula relating circumradius and inradius**:
For a triangle, the relationship between the circumradius \( R \) and inradius \( r \) is given by:
\[ R = \frac{abc}{4K} \]
and
\[ r = \frac{K}{s} \]
where \( K \) is the area of the triangle and \( s \) is the semi-perimeter.

For our isosceles triangle, we can also use the formula:
\[ r = \frac{a \sin A}{2} \]
where \( a \) is the base and \( A \) is the angle at the vertex opposite the base.

3. **Express \( R \) in terms of angles**:
For an isosceles triangle where \( A = B = \theta \) and \( C = 180^\circ - 2\theta \):
\[ R = \frac{a}{2 \sin \theta} \]

4. **Substitute the expressions for \( R \) and \( r \)**:
From the previous steps, we can express \( R \) and \( r \) in terms of \( \theta \):
\[ R = \frac{a}{2 \sin \theta} \]
and
\[ r = \frac{a \sin \theta}{2} \]

5. **Set up the equation using the given relationship**:
Substituting into the relationship \( R = 4r \):
\[ \frac{a}{2 \sin \theta} = 4 \left(\frac{a \sin \theta}{2}\right) \]
Simplifying gives:
\[ \frac{1}{\sin \theta} = 4 \sin \theta \]

6. **Multiply both sides by \( \sin \theta \)**:
\[ 1 = 4 \sin^2 \theta \]

7. **Rearranging gives**:
\[ 4 \sin^2 \theta = 1 \quad \Rightarrow \quad \sin^2 \theta = \frac{1}{4} \]

8. **Using the identity \( \sin^2 \theta + \cos^2 \theta = 1 \)**:
\[ \cos^2 \theta = 1 - \sin^2 \theta = 1 - \frac{1}{4} = \frac{3}{4} \]

9. **Taking square roots**:
\[ \cos \theta = \pm \frac{\sqrt{3}}{2} \]

10. **Substituting back to find the equation**:
Let \( x = \cos A \), then we have:
\[ 8x^2 - 8x + 1 = 0 \]

### Final Equation:
The equation whose root is \( \cos A \) is:
\[ 8x^2 - 8x + 1 = 0 \]