Suppose AC = 5 cm, BC = 12 cm, and mAC = 45.2°.
To the nearest tenth of a unit, the radius of the circumscribed circle
is __?__ cm and m∠OAC = __?__°.


Help pls

Suppose AC 5 cm BC 12 cm and mAC 452 To the nearest tenth of a unit the radius of the circumscribed circle is cm and mOAC Help pls class=

Respuesta :

9514 1404 393

Answer:

  • 6.5 cm
  • 67.4°

Step-by-step explanation:

AB is shown as a diameter, which means that inscribed angle C is a right angle. The length of the diameter can be found from the Pythagorean theorem to be ...

  AB^2 = AC^2 + BC^2

  AB = √(5^2 +12^2) = 13

The radius is half the length of the diameter, so is ...

  OA = 13 cm/2 = 6.5 cm . . . . radius

__

The measure of inscribed angle B is half the measure of arc AC, so is ...

  ∠B = 45.2°/2 = 22.6°

The measure of angle OAC is the complement of this, so is ...

  ∠OAC = 90° -22.6°

  ∠OAC = 67.4°

Answer:

Step-by-step explanation:

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