Points $\mathrm{A}$ and $\mathrm{B}$ are on a circle with centre $\mathrm{O}$. Point $\mathrm{C}$ is on the major arc $\mathrm{AB}$. If $\angle \mathrm{OAC}=35^{\circ}$ and $\angle \mathrm{OBC}=45^{\circ}$, then what is the measure (in degrees) of the angle subtended by the minor arc $\mathrm{AB}$ at the centre?

160

In \(\Delta \)OAC

= OA = OC (radius)

So, \(\angle\)OAC = \(\angle\)OCA = \({35}^\circ\)

Also, In \(\Delta \)OBC

= OB = OC (radius)

So, \(\angle\)OBC = \(\angle\)OCB = \({45}^\circ\)

= \(\angle\)ACB =  \(\angle\)OCA + \(\angle\)OCB = 35 + 45 = 80

We know that angle subtended by an arc on the center of the circle is twice the angle subtended by it on any other part of the circle.

= \(\angle\)AOB = 2 x \(\angle\)ACB = 2 x 80 = \({160}^\circ\)

Therefore, answer is \({160}^\circ\).