What is radius of circumscribed circle to triangle ABC if AC = square root 2, angle B= 45?

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You should know that by the trigonometric law of sine, you may evaluate the radius of circumscribed angle such that:

`(BC)/sinA = (AB)/sin C = (AC)/sinB = 2R`

Notice that the problem provides the length of side AC and the measure of angle B, hence, you should substitute `sqrt2` for AC and `45^o` for B such that:

`(sqrt2)/(sin45^o) = 2R`

You should remember that `sin 45^o = sqrt2/2` such that:

`(sqrt2)/(sqrt2/2) = 2R =gt 1/(1/2)= 2R =gt R = 1`

**Hence, evaluating the radius of circumscribed circle yields R=1.**

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