in figure 3.10 line ab parallel line de find the measures of angle Dr B and Angle A R using given measures of some angles
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Since AB || DE,
∠EDA and ∠BAD are alternate interior angles.
⇒ ∠EDA = ∠BAD = 70°
and , ∠EDR = ∠EDA = 70°
now, in the ΔRED ,
∠DRE + ∠RED + ∠RDE = 180° [ angle sum property of a triangle ]
∴ ∠DRE + 40° + 70° = 180°
⇒ ∠DRE + 130° = 180°
⇒ ∠DRE = 180° - 130°
⇒ ∠DRE = 50°
now,
∠DRE + ∠ARE = 180° [ since AD is a straight line ]
⇒ 50° + ∠ARE = 180°
⇒ ∠ARE = 180° - 50°
⇒ ∠ARE = 130°
alternate method:
∠ARE is an exterior angle for ΔRED
so ∠RED + ∠RDE = ∠ARE
⇒ 70° + 40° = 130° = ∠ARE
∠EDA and ∠BAD are alternate interior angles.
⇒ ∠EDA = ∠BAD = 70°
and , ∠EDR = ∠EDA = 70°
now, in the ΔRED ,
∠DRE + ∠RED + ∠RDE = 180° [ angle sum property of a triangle ]
∴ ∠DRE + 40° + 70° = 180°
⇒ ∠DRE + 130° = 180°
⇒ ∠DRE = 180° - 130°
⇒ ∠DRE = 50°
now,
∠DRE + ∠ARE = 180° [ since AD is a straight line ]
⇒ 50° + ∠ARE = 180°
⇒ ∠ARE = 180° - 50°
⇒ ∠ARE = 130°
alternate method:
∠ARE is an exterior angle for ΔRED
so ∠RED + ∠RDE = ∠ARE
⇒ 70° + 40° = 130° = ∠ARE
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2
Hope this ans helps you
Since AB || DE,
∠EDA and ∠BAD are alternate interior angles.
⇒ ∠EDA = ∠BAD = 70°
and , ∠EDR = ∠EDA = 70°
now, in the ΔRED ,
∠DRE + ∠RED + ∠RDE = 180° [ angle sum property of a triangle ]
∴ ∠DRE + 40° + 70° = 180°
⇒ ∠DRE + 130° = 180°
⇒ ∠DRE = 180° - 130°
⇒ ∠DRE = 50°
now,
∠DRE + ∠ARE = 180° [ since AD is a straight line ]
⇒ 50° + ∠ARE = 180°
⇒ ∠ARE = 180° - 50°
⇒ ∠ARE = 130°
alternate method:
∠ARE is an exterior angle for ΔRED
so ∠RED + ∠RDE = ∠ARE
⇒ 70° + 40°
= 130° = ∠ARE
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