In figure, lines AB and CD intersect at O. If ∠AOC + ∠BOE = 70° and ∠BOD = 40°, find ∠BOE and reflex ∠COE.

∵ Lines AB and CD intersect at O
∴ ∠AOC = ∠BOD
| Vertically Opposite Angles
But ∠BOD = 40°    ...(1) | Given
∴ ∠AOC = 40°    ...(2)
Now, ∠AOC + ∠BOE = 70°
⇒ 40° + ∠BOE = 70° | Using (2)
⇒ ∠BOE = 70° - 40°
⇒ ∠BOE = 30°
Again,
Reflex ∠COE
= ∠COD + ∠BOD + ∠BOE
= ∠COD + 40° + 30°
| Using (1) and (2)
= 180° + 40° + 30°
| ∵ Ray OA stands on line CD
|∴ ∠AOC + ∠AOD = 180° (Linear Pair Axiom) ⇒ ∠COD = 180°
= 250°.

x + y = w + z    ...(1) | Given
∵ The sum of all the angles round a point is equal to 360°.
 x + y + w + z = 360⁰  
 x + y + x + y = 360⁰                       | Using (1)
   2(x + y) = 360⁰


           x + y = 180⁰

∴ AOB is a line.

| If the sum of two adjacent angles is 180°, then the non-common arms of the angles form a line.

∵ Ray OR is perpendicular to line PQ.
∴ ∠QOR = ∠POR = 90°    ...(1)
∠QOS = ∠QOR + ∠ROS    ...(2)
∠POS = ∠POR - ∠ROS    ...(3)
From (2) and (3),
∴ ∠QOS - ∠POS = (∠QOR - ∠POR) + 2∠ROS = 2∠ROS | Using (1)

∵ Ray QP stands on line ST
∴ ∠PQS + ∠PQR = 180°    ...(1)
| Linear Pair Axiom
∵ Ray RP stands on line ST
∴ ∠PRQ + ∠PRT = 180°    ...(2)
| Linear Pair Axiom
From (1) and (2), we obtain
∠PQS + ∠PQR = ∠PRQ + ∠PRT
⇒    ∠PQS = ∠PRT.
| ∵ ∠PQR = ∠PRQ (Given)