$$O P = V _ { 1 } ^ { 2 } + V _ { 2 } ^ { 2 } + 2 \left( V _ { 1 } \right) \left( V _ { 2 } \right) \cos \theta$$ The diameter of a number 10 AGW wire is 0.0285 inch. What is the radius in centimeters? Answer in three decimal places.

Name: \[O P = V _ { 1 } ^ { 2 } + V _ { 2 } ^ { 2 } + 2 \left( V _ { 1 } \right) \left( V _ { 2 } \right) \cos \theta\] The diameter of a number 10 AGW wire is 0.0285 inch. What is the radius in centimeters? Answer in three decimal places.
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The Answer is .036

OP=V12+V22+2(V1)(V2)cos⁡θO P = V _ { 1 } ^ { 2 } + V _ { 2 } ^ { 2 } + 2 \left( V _ { 1 } \right) \left( V _ { 2 } \right) \cos \thetaOP=V12​+V22​+2(V1​)(V2​)cosθ

$O P = V _ { 1 } ^ { 2 } + V _ { 2 } ^ { 2 } + 2 \left( V _ { 1 } \right) \left( V _ { 2 } \right) \cos \theta$