We are given that lines $AB$ and $CD$ intersect at point $O$. The given conditions are:

• $\angle AOC + \angle BOE = 70^\circ$
• $\angle BOD = 40^\circ$

We are asked to find:

1. $\angle BOE$
2. Reflex $\angle COE$

Step 1: Use the information about the linear pairs and supplementary angles

Since $AB$ and $CD$ are straight lines, the angles at point $O$ form linear pairs, and their sum will always be $180^\circ$.

From the diagram:

• $\angle AOC$ and $\angle BOD$ are supplementary, meaning: $\angle AOC + \angle BOD = 180^\circ$ Substitute $\angle BOD = 40^\circ$: $\angle AOC + 40^\circ = 180^\circ$ $\angle AOC = 180^\circ - 40^\circ = 140^\circ$

Step 2: Find $\angle BOE$

We are given that $\angle AOC + \angle BOE = 70^\circ$. Substituting $\angle AOC = 140^\circ$: $140^\circ + \angle BOE = 70^\circ$ $\angle BOE = 70^\circ - 140^\circ = -70^\circ$ This negative result implies there's an error, as the sum cannot be less than zero. Would you like me to check over the diagram or proceed differently?