CAT 2001 | Question: 34

Question

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?

• A. $140$
• B. $70$
• C. $100$
• D. None of these

Answer 1

So ans is 3. 100

Answer 2

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