Question

3. In the given figure, if Z CDE = ZDEC,
then 2 AED = ?
(a) 124
(6) 112°
(c) 68°
(d) 120°
​

Answer 1

Answer:

a)124

Step-by-step explanation:

by exterior angle theorem

angle ACD=68

CDE=DEC=56

Hence AED=ECD+CDE=68+56=124

Answer 2

Answer:

(a) 124°

Step-by-step explanation:

In Δ ABC,

<BAC + <ABC = <DCE

{by exterior angle property of Δ}

48° + 20° = <DCE

<DCE = 68°

<CDE = <DEC {given}

Let <CDE = <DEC = x

Now, Δ CDE is an isosceles Δ

So by angle sum property of Δ,

x + x + <DCE = 180°

2x = 180° - 68°

x = 112°/2

x = 56°

By exterior angle property,

<AED = <DCE + <CDE

<AED = 68° + 56°

<AED = 124°

Therefore, the answer is (a) 124°.

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