Question

In $\Delta ABC,\angle A={30}^{\circ },\angle B={40}^{\circ }and\angle C={110}^{\circ }$. 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. ${60}^{\circ },{40}^{\circ },{80}^{\circ }$
• C. ${30}^{\circ },{40}^{\circ },{110}^{\circ }$
• D. ${60}^{\circ },{70}^{\circ },{50}^{\circ }$

Answer 1

The correct option is C

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

In $\Delta ABC$,

$\angle A={30}^{\circ },\angle B={40}^{\circ },\angle C={110}^{\circ }$

D, E and F are mid -points of the sides of the triangle . By joining them in order, DEF is a triangle formed

Now BDEF, CDFE and AFDE are || gm

$\therefore \angle A=\angle D={30}^{\circ }$

$\angle B=\angle E={40}^{\circ }$

$\angle C=\angle F={110}^{\circ }$

$\therefore$ Angles are ${30}^{\circ },{40}^{\circ },{110}^{\circ }$