I bymistake, sent 2 same questions, but in the 1st question itself I was told that it was missing the value of one angle by Neha Sethi ma'am.
I rechecked it and posted it again as I was asked.
But, then also I did not get the answer.


This time please answer it here or in the old ones. 
​Please Neha Sethi ma'am.
 

Dear Student,


since OA bisects BOGlet BOG= 2ythen BOA=AOG=ysimilarly let COE=2xthen COD=DOW=xsince EOG is straight lineFOG+FOE=1803×2y+70=1806y=110y=1106y=553° Also BOG+BOC+COE=1802y+x+90+2x=1802y+3x=901103+3x=903x=90-1103x=30-1109x=270-1109x=1609hence value  of angles areAOG=AOB=y=553° BOC=x+90=17.77+90=107.77COD=DOE=x=1609FOG=6y=110FOE=70

Regards,

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