In the figure given below, what will be the measures of $\angle \mathrm{AOC},\angle \mathrm{COF},\angle \mathrm{DOE},$$\text{and }\angle \mathrm{BOF}$ If the lines AB, CD, and EF intersect at O?

It is given that the lines AB, CD, and EF intersect at O.

Using properties of the vertical angles, we have:
$\begin{array}{l}\angle \mathrm{AOE}=\mathbf{\angle }\mathbf{BOF}={\mathbf{35}}^{\mathbf{°}}\text{ (1)}\\ \mathbf{\angle }\mathbf{AOC}=\angle \mathrm{BOD}={\mathbf{40}}^{\mathbf{°}}\text{ (2)}\end{array}$

Using linear pair property, we have:
$\angle \mathrm{AOE}+\angle \mathrm{EOB}={180}^{°}$ 

 $\Rightarrow \angle \mathrm{AOE}+\angle \mathrm{DOE}+\angle \mathrm{BOD}={180}^{°}$
$\begin{array}{l}\Rightarrow {35}^{°}+\angle \mathrm{DOE}+{40}^{°}={180}^{°}\\ \Rightarrow \mathbf{\angle }\mathbf{DOE}\mathbf{=}{\mathbf{105}}^{\mathbf{°}}\end{array}$
 

Using properties of the vertical angles,

  $\begin{array}{l}\angle \mathrm{DOE}=\angle \mathrm{COF}\\ \Rightarrow \mathbf{\angle }\mathbf{COF}\mathbf{ }\mathbf{=}\mathbf{ }{\mathbf{105}}^{\mathbf{°}}\end{array}$

Therefore, we have got:
$\mathbf{\angle }\mathbf{AOC}\mathbf{=}{\mathbf{40}}^{\mathbf{°}}\mathbf{,}\mathbf{\angle }\mathbf{COF}\mathbf{=}{\mathbf{105}}^{\mathbf{°}}\mathbf{,}\mathbf{\angle }\mathbf{DOE}\mathbf{=}{\mathbf{105}}^{\mathbf{°}}\mathbf{,}\mathbf{\angle }\mathbf{BOF}\mathbf{=}{\mathbf{35}}^{\mathbf{°}}\mathbf{.}$