If ac = 5 cm, bc = 12 cm, and m archac =40 , the radius of the circumscribed circle and m∠oac =°.

radius = 6.5

<OAC = 67.38 degrees

Step-by-step 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.

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

Step-by-step explanation:

the expected value of the random variable : 0, 1 and 2

step-by-step explanation:

∠lno ≅ ∠lnm

∠oln ≅ ∠mln

step-by-step explanation:

given   that triangle lno is congruent to triangle lnm.

we have to fill up the blanks in the given question.

according to above order of congruency   kept in mind we can say that

∠lno ≅ ∠lnm

∠oln ≅ ∠mln

ln ≅ ln due to common in both.

hence, by sas congruency,

the side angle side rule   states that if two sides and the included angle of one triangle are congruent to another triangle, then these two triangles are congruent.