$A,B,C,D$ lie on circle such that chord $BC$ subtend ${110}^{o}$ at centre $O$ of circle, then $\angle{BEC}$ is

By the central angle theorem, the central angle is twice the inscribed angle when both subtend the same arc, so angle BAC is 110 divided by 2, or 55 degrees. In triangle ABE, the external angle BEC equals the sum of the two opposite internal angles: angle BEC equals angle EAB plus angle EBA. Therefore, angle BEC equals 55 degrees minus 15 degrees, resulting in 40 degrees.

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