Question

AB is parallel to CD.angle ABC=65°, angle CDE=15° and AB =AE what is the value of angle AEF​

Answer 1

Answer:

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Step-by-step explanation:

∠ABC = 65° and ∠CDE = 15°

Here, ∠ABC + ∠TCB = 180°

(∵ AB || CD)

∠TCB = 180° – ∠ABC

∴ ∠TCB = 180° – 65° = 115°

∵ ∠TCB + ∠DCB = 180°

(Linear pair)

∴ ∠DCB = 65°

Now, in ΔCDE

∠CED = 180° – (∠ECD + ∠EDC)

(∵ ∠ECD = ∠BCD)

= 180° – (– 65° + 15° ) = 100°

∵ ∠DEC + ∠FEC = 180°

⇒ ∠FEC = 180° – 100° = 80°

Given that, AB = AE.

i.e. ΔABE an isosceles triangle.

∴ ∠ABE = ∠AEB = 65°

∵ ∠AEB + ∠AEF + ∠FEC = 180°

(straight line)

⇒ 65° + x° + 80° = 180°

∴ x° = 180° – 145° = 35°.

Answer = 35°

Answer 2

Answer:

Step-by-step explanation: