Math Problem Statement

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?

Solution

Thank you for providing additional details! Now it's clear that point CC is on the circle centered at OO with radius AO=CO=6AO = CO = 6. Let's solve for the length of CBCB.

Given:

  • AO=6AO = 6 (radius of the circle)
  • CO=6CO = 6 (radius of the circle, same as AOAO)
  • OAB=90\angle OAB = 90^\circ (right angle at AA)
  • AB=10AB = 10

Step-by-Step Solution:

  1. Understanding the Geometry:

    • Triangle OABOAB is a right triangle with OAB=90\angle OAB = 90^\circ.
    • Points AA and CC lie on a circle centered at OO with radius 66.
  2. Position of Point CC:

    • Since CC is another point on the circle, CO=AO=6CO = AO = 6.
  3. Finding OB:

    • Since OAB\triangle OAB is a right triangle, use the Pythagorean theorem to find OBOB: OB2=OA2+AB2OB^2 = OA^2 + AB^2 Plugging in the known values: OB2=62+102OB^2 = 6^2 + 10^2 OB2=36+100OB^2 = 36 + 100 OB2=136OB^2 = 136 OB=136=234OB = \sqrt{136} = 2\sqrt{34}
  4. Using Circle Geometry:
    Since AA, BB, and CC form a cyclic quadrilateral (all points lie on the circle), the opposite angles of a cyclic quadrilateral sum to 180180^\circ. However, in this case, OAC\triangle OAC and OBC\triangle OBC share the radius CO=6CO = 6, making them both isosceles triangles.

  5. Calculate the Length of CBCB:

    • Points CC and AA lie on the circle. Since the circle's radius is 66 and points CC and AA are on the circumference, we can use the fact that triangle OCBOCB is isosceles with CO=OB=6CO = OB = 6.

    To find CBCB, we need the coordinates for a more precise calculation. However, we can observe that in the circle, with OB=234OB = 2\sqrt{34} and the law of cosines in triangle OCBOCB: CB2=CO2+OB22COOBcos(θ),CB^2 = CO^2 + OB^2 - 2 \cdot CO \cdot OB \cdot \cos(\theta), where θ\theta is the angle COB\angle COB.

  6. Finding the Angle θ=COB\theta = \angle COB:

    • Since triangle OABOAB is a right triangle and inscribed in the circle, points AA and BB are endpoints of a diameter. Thus, OCB\triangle OCB has a right angle OCB=90\angle OCB = 90^\circ.
  7. Apply the Pythagorean Theorem to OCB\triangle OCB: CB2=CO2+OB2CB^2 = CO^2 + OB^2 CB2=62+(234)2CB^2 = 6^2 + (2\sqrt{34})^2 CB2=36+4×34CB^2 = 36 + 4 \times 34 CB2=36+136CB^2 = 36 + 136 CB2=172CB^2 = 172 CB=172=243CB = \sqrt{172} = 2\sqrt{43}

Conclusion:

The length of CBCB is 2432\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 OABOAB?
  2. How would you find the circumference of the circle with center OO and radius 66?
  3. What is the equation of the circle with center OO and radius 66?
  4. Can you find the coordinates of points AA and BB given their positions relative to OO?
  5. If a tangent is drawn from point BB 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.

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Math Problem Analysis

Mathematical Concepts

Geometry
Trigonometry
Circle Geometry

Formulas

Pythagorean theorem
Law of cosines

Theorems

Properties of cyclic quadrilaterals

Suitable Grade Level

High School