Question

in the given figure BE is parallel to CD , angle abc is equal to 120 degrees , angle C is equal to 80 degrees , angle CDE is equal to 110 degrees and BC is equal to CD . find angle BAE , angle CBE and angle BDE.

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Answer 1

Answer:

Angle BAE=50°, CBE=100° and BDE=50°

Answer 2

Answer :-

• ∠BAE = 50°
• ∠CBE = 100°
• ∠BDE = 60°

Given :-

• ∠ABC = 120°
• ∠C = 80°
• ∠CDE = 110°

To Find :-

• ∠BAE = ?
• ∠CBE = ?
• ∠BDE = ?

Solution :-

✯ Let's first find ∠BAE.

◆ In quadrilateral ABCD,

⇒ ∠A + ∠B + ∠C + ∠D = 360°  [Sum of all the angles of quadrilateral is 360°]

⇒ ∠BAE + ∠ABC + ∠C + ∠CDE = 360°

⇒ ∠BAE + 120° + 80° + 110° = 360°

⇒ ∠BAE + 310° = 360°

⇒ ∠BAE = 360° - 310°

⇒ ∠BAE =  50°

∴ ∠BAE = 50°

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✯ Now let's find ∠CBE.

◆ In △BCD,

⇒ ∠DBC + ∠C + ∠CDB = 180°  [Sum of all the angles of triangle is 180°]

⇒ 2∠DBC +∠C = 180°  [△BCD is an isosceles triangle so ∠DBC = ∠CDB]

⇒ 2∠DBC + 80° = 180°

⇒ 2∠DBC = 180° - 80°

⇒ 2∠DBC = 100°

⇒ ∠DBC = $\sf \dfrac{100^\circ}{2}$

⇒ ∠DBC = 50°

∴ ∠DBC = 50°

∴ ∠CDB = 50°

◆ Now we are given that DE || DC

⇒ ∠DEB + ∠CDE = 180°  [Sum of co-interior angles is 180°]

⇒ ∠DEB + 110° = 180°

⇒ ∠DEB = 180° - 110°

⇒ ∠DEB = 70°

∴ ∠DEB = 70°

◆ In quadrilateral BCDE,

⇒ ∠B + ∠C + ∠D + ∠E = 360°  [Sum of all the angles of quadrilateral is 360°]

⇒ ∠CBE+ ∠C + ∠CDE + ∠DEB = 360°

⇒ ∠CBE + 80° + 110° + 70° = 360°

⇒ ∠CBE + 260° = 360°

⇒ ∠CBE = 360° - 260°

⇒ ∠CBE =  100°

∴ ∠CBE = 100°

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✯ Now let's find ∠BDE.

◆ ∠BDE + ∠CDB = ∠CDE

⇒ ∠BDE + 50° = 110°

⇒ ∠BDE = 110° - 50°

⇒ ∠BDE = 60°

∴ ∠BDE = 60°

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