Question

please solve this question

Answer 1

In this figure,

$\angle {AOC} = {BOD}$ [Opposite Angle]

=>$\angle{AOC}$ = 35°

$\angle {BOF} = {AOE}$

=>$\angle{BOF}$ = 40°


Now,
$\angle{AOE} + {DOE} + {BOD}$ = 180°

=> 40° + $\angle{DOE}$ + 35° = 180°

=> $\angle {DOE} = 180° - (40°+35°) =>{DOE}$ = 105°

•°• $\angle{COF} = {DOE}$

=> $\angle {COF}$ = 105°


$\ boxed{ANSWER}$ ➡

[tex]\angle {AOC} = 35°

{COF} = 105°

{DOE} = 105°

{BOF} = 40°

Answer 2

Given,

$\angle \: AOE = 40° \\ \angle \: BOD = 35° \\ \\ Now, \\ \angle \: AOE \: + \angle \: BOD + \angle \: EOD = 180° \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: ... \: (linear \: pair) \\ \\ 40° + 35° + \angle \: EOD = 180° \\ \\ \angle \: EOD = 180° - 75" \\ \\ \bf \: \angle \: EOD = 105°$
$\angle \: EOD = \angle \: COF \\ \: \: \: \: \: \: \: \: \: \: \: ... (vertically \: opposite \: angle) \\ \\ \bf \: \angle \: COF = 105°$
$Similarly, \\ \angle \: AOE = \angle \: BOF \\ \angle \: BOD = \angle \: COA \\ \\ \bf \: \angle \: BOF = 40° \\ \bf \: \angle \: COA = 35°$

If any doubt, please ask :)