Question

In the adjacent figure, ABC is a triangle in which $\angle B={50}^o$ and $\angle C={70}^o$. Sides AB and AC are produced. If 'z' is the measure of the angle between the bisectors of the exterior angles of formed, then find 'z'.

Answer 1

$x^{\circ }+x^{\circ }+{50}^{\circ }={180}^{\circ }$ (Straight angle made on $AB$ at point $B$)

$2x+{50}^{\circ }={180}^{\circ }2x={180}^{\circ }-{50}^{\circ }={130}^{\circ }x=\frac{{130}^{\circ }}{2}={65}^{\circ }$

$y^{\circ }+y^{\circ }+{70}^{\circ }={180}^{\circ }$ (Straight angle made on $AC$ at point $C$)

$2y+{70}^{\circ }={180}^{\circ }2y={180}^{\circ }-{70}^{\circ }={110}^{\circ }y=\frac{{110}^{\circ }}{2}={55}^{\circ }$

In $△BOC$

$x+y+z={180}^{\circ }$ (Angle sum property)

${65}^{\circ }+{55}^{\circ }+z={180}^{\circ }z={180}^{\circ }-{120}^{\circ }={60}^{\circ }$