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Re: If ∠ABC = 150° and ∆CED is isosceles, what is the value of ∠ [#permalink]
1
should not it be 90+90+150+ E= 360[/quote]

∠A + ∠B + ∠D + ∠E = 360° --(equation 1)

∠A = ∠E = 90° (represented in fig)
∠B = 150° (given in que)
when you put all these three angular values in equation 1 :
∠A + ∠B + ∠D + ∠E = 360°
90° + 150° + ∠D + 90° = 360°
∠D = 30°
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Re: If ∠ABC = 150° and ∆CED is isosceles, what is the value of ∠ [#permalink]
vndnjn wrote:
In quadrilateral, ABCE: sum of all angles = 360°
90° + 90° + 150° + ∠D = 360°
∠D = 30°
∆CED is isosceles, CE = CD and ∠E = ∠D = 30°


You said CED is isosceles, so CE = CD. How does one reach this conclusion? Can't ED=CD?

The problem I'm having here is that ∆CDE, while isosceles can be 30-30-120 or 75-75-30.

This means our desired angle ∠CED will either be 30 or 75.

Please tell me where I am going wrong. Thank you!
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Re: If ABC = 150° and CED is isosceles, what is the value of [#permalink]
i think this question is vague and conjectural there isn't specified that the triangle is isosceles how? which two sides are equal please correct this.
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Re: If ABC = 150° and CED is isosceles, what is the value of [#permalink]
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