In , = 30^ , = 40^ and = 110^ . The angles of the triangle formed…

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:

✓
$30^\circ , 40^\circ , 110^\circ $
B
$70^\circ , 70^\circ , 40^\circ$
C
$60^\circ , 40^\circ , 80^\circ$
D
$60^\circ , 70^\circ , 50^\circ$

✓

Answer

Correct option: A.

$30^\circ , 40^\circ , 110^\circ $

Given,
In $\angle\text{ABC}$

$\angle\text{A} = 30^\circ,\ \angle\text{B} = 40^\circ,\ \angle\text{C} = 110^\circ$
In figure, $BDEF, DCEF, DEAF$ are parallelograms so,
$\angle\text{B} = \angle\text{E} = 40^\circ [∵ \angle\text{B}$ & $\angle\text{E}$ are opposite angles of parallelogram $BDEF]$
$\angle\text{C} = \angle\text{F} = 110^\circ [∵ \angle\text{C}$ & $\angle\text{F}$ are opposite angles of parallelogram $DCEF]$
$\angle\text{A} = \angle\text{D} = 30^\circ [∵ \angle\text{A}$ & $\angle\text{D}$ are opposite angles of parallelogram $DEAF]$
Hence, $\angle\text{D} = 30^\circ$
$\angle\text{E} = 40^\circ$
$\angle\text{F} = 110^\circ$