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 for cos A given that the circumradius \( R \) of isosceles triangle \( \Delta ABC \) is 4 times its inradius \( r \). We know that \( \angle A = \angle B \) and \( BC = AC \).

### Step-by-Step Solution:

1. **Understanding the relationship between circumradius and inradius:**
Given that \( R = 4r \), we can express this relationship in terms of the angles of the triangle.

2. **Using the formula for circumradius and inradius:**
The circumradius \( R \) of a triangle is given by:
\[ R = \frac{abc}{4K} \]
where \( K \) is the area of the triangle. The inradius \( r \) is given by:
\[ r = \frac{K}{s} \]
where \( s \) is the semi-perimeter of the triangle.

3. **Setting up the triangle:**
Since \( \Delta ABC \) is isosceles with \( \angle A = \angle B \), let \( \angle A = \angle B = x \) and \( \angle C = 180^\circ - 2x \).

4. **Expressing the circumradius and inradius in terms of angles:**
For an isosceles triangle, we can use the formula for \( R \) and \( r \):
\[ R = \frac{a}{2 \sin A} \]
and
\[ r = \frac{K}{s} \]
where \( K \) can be expressed in terms of the sides and angles.

5. **Using the sine rule:**
We can express the area \( K \) in terms of \( R \) and the angles:
\[ K = \frac{1}{2}ab \sin C = \frac{1}{2}a^2 \sin(180^\circ - 2x) = \frac{1}{2}a^2 \sin(2x) \]

6. **Finding the relationship between \( R \) and \( r \):**
From the relationship \( R = 4r \), we can substitute the expressions for \( R \) and \( r \):
\[ \frac{a}{2 \sin x} = 4 \cdot \frac{K}{s} \]

7. **Using the sine double angle identity:**
We know that \( \sin(2x) = 2 \sin x \cos x \). Thus, we can express the area \( K \) as:
\[ K = \frac{1}{2} a^2 (2 \sin x \cos x) = a^2 \sin x \cos x \]

8. **Setting up the equation:**
Now substituting back into the equation, we get:
\[ \frac{a}{2 \sin x} = 4 \cdot \frac{a^2 \sin x \cos x}{s} \]
Simplifying this gives us a quadratic equation in terms of \( \cos x \).

9. **Final equation:**
After simplification, we arrive at the quadratic equation:
\[ 8 \cos^2 x - 8 \cos x + 1 = 0 \]
This can be solved using the quadratic formula.