Question

In the adjacent figure, ABC is a triangle in which $\angle B={50}^{\circ }$ and $\angle C={70}^{\circ }$. Sides AB and AC are produced. If 'z' is the measured of the angle between the bisectors of the exterior angles so formed, then $\angle$x $\angle$y $\angle$z are respectively :

• A. ${65}^o$ ${55}^o$ ${60}^o$
• B. ${55}^o$ ${45}^o$ ${35}^o$
• C. ${60}^o$ ${65}^o$ ${55}^o$
• D. ${65}^o$ ${45}^o$ ${35}^o$

Answer 1

The correct option is A ${65}^o$ ${55}^o$ ${60}^o$

$AB$ is straight line

$x+x+50=180$

$2x+50=180$

$2x=130$

$x=65$

Also $y+y+70=180$

$2y=110$

$y=55$

Now in $\Delta BCD$

$65+55+z=180$

$\Rightarrow z=180-120$

$z=60$