Dear Student, ∠ACB=∠ECD i vertically opposite anglesIn ∆ABC,∠BAC+∠ABC+∠ACB=180° Angle sum property⇒∠BAC+40°+∠ACB=180° iiIn ∆DEC,∠CDE+∠DEC+∠ECD=180° Angle sum property⇒∠CDE+85°+∠ECD=180° iiiEquating ii and iii we get∠BAC+40°+∠ACB=∠CDE+85°+∠ECD⇒∠BAC+40°=∠CDE+85° by i⇒∠BAC-∠CDE=85°-40°=45° Regards,

Solve question no 27

Dear Student,

∠ A C B = ∠ E C D . . . . . (i) [v e r t i c a l l y o p p o s i t e a n g l e s] I n ∆ A B C, ∠ B A C + ∠ A B C + ∠ A C B = 180 ° [A n g l e s u m p r o p e r t y] \Rightarrow ∠ B A C + 40 ° + ∠ A C B = 180 ° . . . . (i i) I n ∆ D E C, ∠ C D E + ∠ D E C + ∠ E C D = 180 ° [A n g l e s u m p r o p e r t y] \Rightarrow ∠ C D E + 85 ° + ∠ E C D = 180 ° . . . . (i i i) E q u a t i n g (i i) a n d (i i i) w e g e t ∠ B A C + 40 ° + ∠ A C B = ∠ C D E + 85 ° + ∠ E C D \Rightarrow ∠ B A C + 40 ° = ∠ C D E + 85 ° [b y (i)] \Rightarrow ∠ B A C - ∠ C D E = 85 ° - 40 ° = 45 °

Regards,

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