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If AC = 5 cm, BC = 12 cm, and m archAC =40 , the radius of the circumscribed circle is____cm and m∠OAC =____°.

If AC = 5 cm, BC = 12 cm, and m archAC =40 , the radius of the circumscribed circle-example-1

2 Answers

2 votes

Answer:

radius = 6.5

<OAC = 67.38 degrees

Explanation:

It looks like you are assuming 0 is on AB and in fact defines the circumradius of the circle. Further you are assuming (I think) that ABC is a right angle triangle.

Therefore AB^2 = AC^2 + BC^2

AB^2 = 5^2 + 12^2

AB^2 = 25 + 144

AB^2 = 169

AB = 13

AO is therefore 1/2 of 13 = 6.5.

CO is also 6.5 since all radii are equal.

I don't know if you know this but the only way you can solve this is to use the Cosine Law for distance.

AC = 5

OC = 6.5

OA = 6.5

<CAO = ??

CO^2 = OA^2 + CA^2 - 2*OA*CA*Cos(<CAO)

6.5^2 = 6.5^2 + 5^2 - 2*5*6.5*Cos(<CAO)

42.25 = 42.25 + 25 - 65*Cos(<CAO)

0 = +25 - 65*Cos(<CAO)

-25 = -65*Cos(<CAO)

0.3846 = Cos(<CAO)

<CAO = cos-1(0.3846)

<CAO = 67.38 degrees.

User Abhishek
by
5.7k points
0 votes

Answer:

the circle is 6.5 cm and angle OAC= 70 for plato users

Explanation:

User Faisal Khan
by
5.8k points
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