In figure, $BC$ is diameter of the circle and $\angle BAO=60°$. Then $\angle ADC$ is equal to :

In figure, $\angle AOB=90°$ and $\angle ABC=40°$, then $\angle CAO$ is equal to:

In figure, $\mathrm{\angle }\mathrm{A}\mathrm{B}\mathrm{C}={69}^{\mathrm{o}},\mathrm{\angle }\mathrm{A}\mathrm{C}\mathrm{B}={31}^{\mathrm{o}}$, find $\mathrm{\angle }\mathrm{B}\mathrm{D}\mathrm{C}.$

In figure, $\mathrm{A},\mathrm{B},\mathrm{C}$ and $\mathrm{D}$ are points on a circle. $\mathrm{A}\mathrm{C}$ and $\mathrm{B}\mathrm{D}$ intersect at a point $\mathrm{E}$ such that $\mathrm{\angle }\mathrm{B}\mathrm{E}\mathrm{C}={130}^{\mathrm{o}}$ and $\mathrm{\angle }\mathrm{E}\mathrm{C}\mathrm{D}={20}^{\mathrm{o}}.$ Find $\mathrm{\angle }\mathrm{B}\mathrm{A}\mathrm{C}.$