kindly provide the answer Share with your friends Share 0 Varun.Rawat answered this Since, ∠BDE is an exterior ∠ of ∆BDC, then ∠BDE = ∠CBD + ∠BCD Exterior angle theorem⇒∠BDE = 45°+40° = 85°⇒x = 85°In ∆BDE, ∠BDE + ∠BED + ∠DBE = 180° angle sum property⇒85° + 35° + ∠DBE = 180°⇒∠DBE = 180° - 120° = 60°now, ∠CBE = ∠CBD + ∠DBE = 45° + 60° = 105°Now, ∠CBE + ∠ABE = 180° Linear Pair⇒105° + y = 180°⇒y = 180° - 105° = 75° 2 View Full Answer Pawan Sharma answered this X=<CBD + <BCD (some of extirior angle =sum of opp. interior angle) =85 Now, <X+ <DBE+ <BED=180 180 =85 + 35+<BED 120+<BED=180 <BED=60 Then, y + 45 + Y=180 60+45+,Y=180 105+<Y=180 <Y=75 -1
