In a circle with centre $\mathrm{O}$, points $\mathrm{A}, \mathrm{B}, \mathrm{C}$ and $\mathrm{D}$ in this order are concyclic such that $\mathrm{BD}$ is a diameter of the circle. If $\angle B A C=22^{\circ}$, then find the measure (in degrees) of $\angle C O D$.

136

Here, according to the concept,

\(\angle\)BAD = 90

= \(\angle\)BAC + \(\angle\)CAD = 90

= \(\angle\)CAD = 90 - 22 = 68

Now, we know that

\(\angle\)COD = 2 x \(\angle\)CAD

= \(\angle\)COD = 2 x 68 = 136

Therefore, \(\angle\)COD is \({136}^\circ\).