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In $\triangle \text{DEF}$ shown below, points $\text{A, B,}$ and $\text{C}$ are taken on $\text{DE, DF}$ and $\text{EF}$ respectively such that $\text{EC = AC}$ and $\text{CF = BC}.$ If $\measuredangle \text{D} = 40^{\circ}$, then what is $\measuredangle \text{ACB}$ in degrees?

  1. $140$
  2. $70$
  3. $100$
  4. None of these
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So ans is 3. 100

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Ans: (3). 100

solution: Let the angle E be x in triangle(AEC),then angle AEB= 180-2*x. Then in triangle DEF, angle F=180-(40+x).

Now in triangleBCF, angle BCF=2*x-100. NOW,angle ACB= 180-(180-2*x+2*x-100)=100
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