In the adjoining figure, AB , CD and EF intersect at O, if
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Step-by-step explanation:
From the figure,
• BOF = AOE
. BOF = 40°
• ZAOC = 2 BOD • AOC = 35°.
1. [vertically opposite /s]
2. [Given AOE = 40°] 3. [Vertically opposite /s]
4. [given BOD = 35°]
Clearly, AOC + COF + BOF = 180°
35° + COF + 40° = 180°
COF + 75° = 180°
COF=180°-75°
COF = 105°.
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From the figure,
\angle∠ BOF = \angle∠ AOE
\angle∠ BOF = 40°
\angle∠ AOC = \angle∠ BOD
\angle∠ AOC = 35°.
[vertically opposite \angle∠ s]
[Given \angle∠ AOE = 40°]
[Vertically opposite \angle∠ s]
[given \angle∠ BOD = 35°]
Clearly, \angle∠ AOC + \angle∠ COF + \angle∠ BOF = 180°
35° + \angle∠ COF + 40° = 180°
\angle∠ COF + 75° = 180°
\angle∠ COF = 180°-75°
\angle∠ COF = 105°.
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