$O P = V _ { 1 } ^ { 2 } + V _ { 2 } ^ { 2 } + 2 \left( V _ { 1 } \right) \left( V _ { 2 } \right) \cos \theta$ Two voltages ( $E _ { 1 }$ and $E _ { 2 }$ ) are applied to a resister. $$E _ { 1 }$$ is 40 volts at $0 ^ { \circ }$ reference. $E _ { 2 }$ is 60 volts and leads $E _ { 1 }$ by $40 ^ { \circ }$ . Find $E_ R$ to the nearer tenth.

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The Answer of $O P = V _ { 1 } ^ { 2 } + V _ { 2 } ^ { 2 } + 2 \left( V _ { 1 } \right) \left( V _ { 2 } \right) \cos \theta$ Two voltages ( $E _ { 1 }$ and $E _ { 2 }$ ) are applied to a resister. $$E _ { 1 }$$ is 40 volts at $0 ^ { \circ }$ reference. $E _ { 2 }$ is 60 volts and leads $E _ { 1 }$ by $40 ^ { \circ }$ . Find $E_ R$ to the nearer tenth.

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

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