Please solve the question in a neat manner
Dear Student,
First we join B to D.
In triangle BCD, BC=CD.
Thus angles CBD and CDB are equal. Let them be x.
x+x+110=180 Therefore, x=35 is the measure of angles CBD and CDB.
Also, arc BC=arc CD=arc AB=2*35=70.
Angle BAE=120 therefore arc BCDE=2*120=240 degrees.
Arc BCDE=arc BC+arc CD+arc DE= 70+70+arc DE=240
Thus arc DE = 100.
Arc BA+ arc AE+arc ED = 2*110=220
arc AE= 220-70-100=50.
2* angle ABC=arc AE=arc ED+ arc DC
Thus angle ABC = 220/2=110.
Similarly angle CDE can be found as 1/2(70+70+50) = 95.
You could similarly use this method to find the measure of the remaining required angles.
Regards.
First we join B to D.
In triangle BCD, BC=CD.
Thus angles CBD and CDB are equal. Let them be x.
x+x+110=180 Therefore, x=35 is the measure of angles CBD and CDB.
Also, arc BC=arc CD=arc AB=2*35=70.
Angle BAE=120 therefore arc BCDE=2*120=240 degrees.
Arc BCDE=arc BC+arc CD+arc DE= 70+70+arc DE=240
Thus arc DE = 100.
Arc BA+ arc AE+arc ED = 2*110=220
arc AE= 220-70-100=50.
2* angle ABC=arc AE=arc ED+ arc DC
Thus angle ABC = 220/2=110.
Similarly angle CDE can be found as 1/2(70+70+50) = 95.
You could similarly use this method to find the measure of the remaining required angles.
Regards.