Calculate ∠AOC ,∠BOD and ∠AOE in the adjoining figure it is bein

1.	Calculate ∠AOC ,∠BOD and ∠AOE in the adjoining figure it is being given that ∠COD=90°,∠BOE=72° and AOB is a straight line.
Answer» ━━━━━━━━━━━━━━━━━━━━━━━━━ $\huge{\underline{\underline{\sf{\orange{Question:-}}}}}$ Calculate ∠AOC ,∠BOD and ∠AOE in the adjoining figure it is being given that ∠COD=90°,∠BOE=72° and AOB is a straight line. ━━━━━━━━━━━━━━━━━━━━━━━━━ $\huge{\underline{\underline{\sf{\orange{Solution:-}}}}}$ Since AOB is a straight line, the SUM of all the angles on the lower SIDE of AOB at a POINT O on it, is 180°. ∴ ∠AOE + ∠BOE = 180° $⟹<klux>3X</klux> + 72 = 180$ $⟹ 3x = (180 - 72) = 108 ⟹ x = \frac{108}{3} = 36. \\$ Again, AOB is a straight line and O is a point on it. So, the sum of all angles on the UPPER side of AOB at a point O on it, is 180°. ∴ ∠AOC + ∠COD + ∠DOB = 180° $= > x + 90 + y = 180$ ⠀[∴ ∠AOC=x°,∠COD=90° and ∠DOB=y°] $⟹ 36 + 9 + y = 180 \: \: \: [\therefore \: x = 36] \\$ $⟹ 126 + y = 180 ⟹ y = (180 - 126) = 54. \\$ ∴ ∠AOC = x° = 36°,∠BOD=y° = 54°, ∠AOE=(3x)° = (3×36)° = 108°. ━━━━━━━━━━━━━━━━━━━━━━━━━