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 $z =$ ?

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

✓

Answer

Correct option: A.

$x + y - 180^\circ$

$\angle\text{ABF}+\angle\text{ABC}=180^\circ \ ($linear pair$)$
$\Rightarrow\text{x}+\angle\text{ABC}=180^\circ$
$\Rightarrow\angle\text{ABC}=180^\circ-\text{x}$
$\angle\text{ACG}+\angle\text{ACB}=180^\circ \ ($linear pair$)$
$\Rightarrow\text{y}+\angle\text{ACB}=180^\circ$
$\Rightarrow\angle\text{ACB}=180^\circ-\text{y}$
In $\triangle\text{ABC},$ by angle sum property
$\angle\text{ABC}+\angle\text{ACB}+\angle\text{BAC}=180^\circ$
$\Rightarrow(180^\circ-\text{x})+(180^\circ-\text{y})+\angle\text{BAC}=180^\circ$
$\Rightarrow\angle\text{BAC}-\text{x}-\text{y}+180^\circ=0$
$\Rightarrow\angle\text{BAC}=\text{x}+\text{y}-180^\circ$
Now, $\angle\text{EAD}=\angle\text{BAC}\ ($vertically opposite angles$)$
$\Rightarrow\text{z}=\text{x}+\text{y}-180^\circ$

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