Question

$\text{In the given figure, two rays BD and CE intersect at a point A . The side BC of}\Delta ABC\text{have been produced on both sided to points F and G respectively .If}\angle ABF=x^{\circ },\angle ACG=y^{\circ }and\angle DAE=z^{\circ }Thenz=?$

$\left(a\right)x+y-180\left(b\right)x+y+180\left(c\right)180-(x+y)\left(d\right)x+y+{360}^{\circ }$

Answer 1

ANSWER:
In the given figure, $\angle ABF+\angle ABC={180}^{\circ }$
(Linear pair of angles)
∴ $x^{\circ }+\angle ABC={180}^{\circ }$
⇒ $\angle ABC={180}^{\circ }-x^{\circ }$
.....(1)
Also, $\angle ACG+\angle ACB={180}^{\circ }$
(Linear pair of angles)
∴ y + $\angle ACB={180}^{\circ }$
⇒ $\angle ACB={180}^{\circ }$− y
.....(2)
Also, $\angle BAC=\angle DAE=z$
.....(3)
(Vertically opposite angles)
In $△ABC$
$\angle BAC+\angle ABC+\angle ACB={180}^{\circ }$
∴ z+ ${180}^{\circ }-x°+{180}^{\circ }-y={180}^{\circ }$
(Angle sum property)
[Using (1), (2) and (3)]
⇒ z = x + y − ${180}^{\circ }$
Hence, the correct answer is option (a).