In the given figure, two rays BD and CE intersect at a point A. The side BC of have been produced on both sides to points F and G respectively. If =x^ , =y^ and =z^ then z = ?

In the given figure, two rays BD and CE intersect at a point A. T…

MCQ

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

A
$x + y + 180$
B
$180 - (x + y)$
✓
$x + y - 180$
D
$x + y + 360^\circ$

✓

Answer

Correct option: C.

$x + y - 180$

In the given figure, $\angle\text{ABF}+\angle\text{ABC}=180^\circ$ (Linear pair of angles)
$\therefore \text{x}^\circ+\angle\text{ABC}=180^\circ$
$\Rightarrow \angle\text{ABC}=180^\circ−\text{x}^\circ ...\ (\text{i})$
Also, $\angle\text{ACG}+\angle\text{ACB}=180^\circ$ (Linear pair of angles)
$\therefore \text{y}′+\angle\text{ACB}=180^\circ$
$\Rightarrow \angle\text{ACB}=180^\circ−\text{y}′\ ...\ \text{(ii)}$
Also, $\angle\text{BAC}=\angle\text{DAE}=\text{z}^\circ \ ....\ \text{(iii)}$ (Vertically opposite angles)
In $\triangle\text{ABC}$
$\angle\text{BAC}+\angle\text{ABC}+\angle\text{ACB}=180^\circ$ (Angle sum property)
$\therefore \text{z}^\circ+180−\text{x}^\circ+180^\circ−\text{y}^\circ=180^\circ$ [Using (1), (2) and (3)]
$\Rightarrow \text{z} = \text{x} + \text{y} – 180$