MCQ
In $\triangle\text{ABC},\angle\text{A}=30^\circ,\angle\text{B}=40^\circ $ and $\angle\text{C}=110^\circ.$ The angles of the triangle formed by joining the mid $-$ points of the sides of this triangle are:
  • A
    $70^\circ , 70^\circ , 40^\circ$
  • B
    $60^\circ , 40^\circ , 80^\circ$
  • $30^\circ , 40^\circ , 110^\circ$
  • D
    $60^\circ , 70^\circ , 50^\circ$

Answer

Correct option: C.
$30^\circ , 40^\circ , 110^\circ$

If in any triangle, all the mid $-$ points $($of each sides$)$ are joined to form a triangle,
then that triangle is similian to a parent triangle.
i.e.$\triangle\text{QPR}\sim\triangle\text{ABC}$
So angles of $\triangle\text{PQR}$ will be same as angles of $\triangle\text{ABC}.$
Thus, angles are $30^\circ , 40^\circ , 110^\circ$

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