Question

plz answer it .don't spam in answer box​

Answer 1

Let's Start:- (i) In triangle BDC

∠DCB + ∠BDC + ∠DBC = 180°                [Angle Sum Property ]

25° + 100° + ∠DBC = 180°

∠DBC = 180° - 125°

∠B = 55°

(ii) ∵ DE ║ BC and DC is transversal

so, ∠EDC = ∠DCB = 55°           [ Alternate interior angles ]

(iii) ∠BDC + ∠EDC + ∠ADE = 180°           [ Linear Pair ]

⇒ 100° + 25° + ∠ADE = 180°

⇒ ∠ADE = 180° - 125°

⇒ ∠ADE = 55°

(iv) In triangle ADE,

∠ADE + ∠DAE + ∠AED = 180°            [Angle Sum Property ]

55° + 55° + ∠AED = 180°

∠AED = 180° - 110°

∠AED = 70°

(v) ∠AED + ∠DEC = 180°             [ Linear Pair ]

70° + ∠DEC = 180°

∠DEC = 180° - 70°

∠DEC = 110°

(vi) In triangle DEC,

∠EDC + ∠DEC + ∠DCE = 180°           [ Angle Sum Property ]

25° + 110° + ∠DCE = 180°

∠DCE = 180° - 135°

∠DCE = 45°

(vii) ∠DEC + ∠DCB = ∠ACB

⇒ 45° + 25° = ∠ACB

⇒ ∠ACB = 70°