This question involves a right triangle and a circle. AO is the radius of the circle. C is another point on the circle, and CO is also the radius. If AO = 6, AB= 10, and the angel of OAB is 90 degrees. What is the length of CB?

Thank you for providing additional details! Now it's clear that point \(C\) is on the circle centered at \(O\) with radius \(AO = CO = 6\). Let's solve for the length of \(CB\).

### Given:
- \(AO = 6\) (radius of the circle)
- \(CO = 6\) (radius of the circle, same as \(AO\))
- \(\angle OAB = 90^\circ\) (right angle at \(A\))
- \(AB = 10\)

### Step-by-Step Solution:

1. **Understanding the Geometry**:  
   - Triangle \(OAB\) is a right triangle with \(\angle OAB = 90^\circ\).
   - Points \(A\) and \(C\) lie on a circle centered at \(O\) with radius \(6\).

2. **Position of Point \(C\)**:
   - Since \(C\) is another point on the circle, \(CO = AO = 6\).

3. **Finding OB**:
   - Since \(\triangle OAB\) is a right triangle, use the Pythagorean theorem to find \(OB\):
   \[
   OB^2 = OA^2 + AB^2
   \]
   Plugging in the known values:
   \[
   OB^2 = 6^2 + 10^2
   \]
   \[
   OB^2 = 36 + 100
   \]
   \[
   OB^2 = 136
   \]
   \[
   OB = \sqrt{136} = 2\sqrt{34}
   \]

4. **Using Circle Geometry**:  
   Since \(A\), \(B\), and \(C\) form a cyclic quadrilateral (all points lie on the circle), the opposite angles of a cyclic quadrilateral sum to \(180^\circ\). However, in this case, \(\triangle OAC\) and \(\triangle OBC\) share the radius \(CO = 6\), making them both isosceles triangles.

5. **Calculate the Length of \(CB\)**:
   - Points \(C\) and \(A\) lie on the circle. Since the circle's radius is \(6\) and points \(C\) and \(A\) are on the circumference, we can use the fact that triangle \(OCB\) is isosceles with \(CO = OB = 6\).

   To find \(CB\), we need the coordinates for a more precise calculation. However, we can observe that in the circle, with \(OB = 2\sqrt{34}\) and the law of cosines in triangle \(OCB\):
   \[
   CB^2 = CO^2 + OB^2 - 2 \cdot CO \cdot OB \cdot \cos(\theta),
   \]
   where \(\theta\) is the angle \(\angle COB\).

6. **Finding the Angle \(\theta = \angle COB\)**:
   - Since triangle \(OAB\) is a right triangle and inscribed in the circle, points \(A\) and \(B\) are endpoints of a diameter. Thus, \(\triangle OCB\) has a right angle \(\angle OCB = 90^\circ\).

7. **Apply the Pythagorean Theorem to \(\triangle OCB\):**
   \[
   CB^2 = CO^2 + OB^2
   \]
   \[
   CB^2 = 6^2 + (2\sqrt{34})^2
   \]
   \[
   CB^2 = 36 + 4 \times 34
   \]
   \[
   CB^2 = 36 + 136
   \]
   \[
   CB^2 = 172
   \]
   \[
   CB = \sqrt{172} = 2\sqrt{43}
   \]

### Conclusion:
The length of \(CB\) is \(2\sqrt{43}\) units.

Would you like more details on any step, or do you have another question?

### Related Questions:
1. What is the area of the right triangle \(OAB\)?
2. How would you find the circumference of the circle with center \(O\) and radius \(6\)?
3. What is the equation of the circle with center \(O\) and radius \(6\)?
4. Can you find the coordinates of points \(A\) and \(B\) given their positions relative to \(O\)?
5. If a tangent is drawn from point \(B\) to the circle, what would be its length?

**Tip**: Remember, when dealing with circles and right triangles, the properties of cyclic quadrilaterals and the Pythagorean theorem are often very useful.

Explore how to find the length of CB in a right triangle and circle problem. Given AO = 6, AB = 10, and a right angle at OAB, discover the step-by-step solution involving circle geometry and the Pythagorean theorem.

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