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⠀ ⠀⠀ ⠀⠀ ⠀⠀ ⋆ ANSWER ⋆
GIVEN : • AB , CD & EF are lines and all intersect at O
• ∠AOC =30° & ∠BOF = 35°
REQUIRED TO FIND : AOE , EOD, BOD & COF
⠀ ⠀ ⠀⠀ ⠀⠀ ⋆ SOLUTION ⋆
FOR ∠AOE ,
➠ here we know that CD , AB & EF intersect each other
so , ∠AOE = ∠FOB ( vertically opposite angle)
( vertically opposite angle) ∴ ∠AOE = 30°
━━━━━━━━━━━━━━━━━━━━
FOR ∠EOD,
➠ here we know that CD is a line
so , ∠COE + ∠EOD = 180° ( angle on a straight line)
[ ∠COE = ∠COA + ∠AOE ]
[∠COA + ∠AOE ] + ∠EOD = 180°
30° + 35° + ∠EOD = 180°
65° + ∠EOD = 180°
∠EOD = 180° - 65°
∴∠EOD = 115°
━━━━━━━━━━━━━━━━━━━━
for ∠BOD ,
➠ here we know that CD , AB & EF intersect each other
so , ∠BOD = ∠AOC ( vertically opposite angle)
∴ ∠BOD = 35°
━━━━━━━━━━━━━━━━━━━━
for ∠COF ,
➠ here we know that CD , AB & EF intersect each other
so , ∠COF = ∠EOD ( vertically opposite angle)
∴∠COF = 115°
━━━━━━━━━━━━━━━━━━━━
ALL ANGLES :-
⋆ ∠AOE = 30°
⋆ ∠EOD = 115°
⋆ ∠BOD = 35°
⋆ ∠COF = 115°
From the figure,
↪∠ BOF = ∠ AOE
⇥∠ BOF = 40°
⇥∠ AOC = ∠ BOD
↪∠ AOC = 35°
[vertically opposite ∠ s]
[Given ∠ AOE = 40°]
[Vertically opposite ∠ s]
[given ∠ BOD = 35°]
Clearly, ∠ AOC + ∠ COF + ∠ BOF = 180°
↛35° + ∠ COF + 40° = 180°
↠∠ COF + 75° = 180°
↠∠ COF = 180°-75°
↠∠ COF = 105°.