Question
In the given figure, if $\text{AB }||\text{ DE}$ and $\text{BD }||\text{ FG}$ such that $\angle\text{FGH}=125^\circ$ and $\angle\text{B}=55^\circ,$ find $x$ and $y.$
✓
Answer
In the given figure, if$\text{AB }||\text{ DE},\text{BD }||\text{ FG},\angle\text{FGH}=125^\circ$ and$\angle\text{B}=55^\circ$
We need to find the value of $x$and $y$
Here, as $AB \| DE$ and $BD$ is the transversal,
so according to the property, "alternate interior angles are equal",
we get $\angle\text{D}=\angle\text{B}$
$\angle\text{D}=55^\circ\dots(1)$ Similarly, as $BD \| FG$ and $DF$ is the transversal $\angle\text{D}=\angle\text{F}$
$\angle\text{F}=55^\circ(\text{Using 1})$ Further, $EGH$ is a straight line.
So, using the property, angles forming a linear pair are supplementary $\angle\text{FGE}=\angle\text{FGH}=180^\circ$
$\text{y}+125^\circ=180^\circ$
$\text{y}=180^\circ-125^\circ$
$\text{y}=55^\circ$ Also, using the property, "an exterior angle of a triangle is equal to the sum of the two opposite interior angles",
we get, In $\triangle\text{EFG}$ with $\angle\text{FGH}$ as its exterior angle $\text{ext.}\angle\text{FGH}=\angle\text{F}+\angle\text{E}$
$125^\circ=55^\circ+\text{x}$
$\text{x}=125^\circ-55^\circ$
$\text{x}=70^\circ$
Thus, $\text{x}=70^\circ$ and $\text{y}=55^\circ$
We need to find the value of $x$and $y$
Here, as $AB \| DE$ and $BD$ is the transversal,
so according to the property, "alternate interior angles are equal",
we get $\angle\text{D}=\angle\text{B}$
$\angle\text{D}=55^\circ\dots(1)$ Similarly, as $BD \| FG$ and $DF$ is the transversal $\angle\text{D}=\angle\text{F}$
$\angle\text{F}=55^\circ(\text{Using 1})$ Further, $EGH$ is a straight line.
So, using the property, angles forming a linear pair are supplementary $\angle\text{FGE}=\angle\text{FGH}=180^\circ$
$\text{y}+125^\circ=180^\circ$
$\text{y}=180^\circ-125^\circ$
$\text{y}=55^\circ$ Also, using the property, "an exterior angle of a triangle is equal to the sum of the two opposite interior angles",
we get, In $\triangle\text{EFG}$ with $\angle\text{FGH}$ as its exterior angle $\text{ext.}\angle\text{FGH}=\angle\text{F}+\angle\text{E}$
$125^\circ=55^\circ+\text{x}$
$\text{x}=125^\circ-55^\circ$
$\text{x}=70^\circ$
Thus, $\text{x}=70^\circ$ and $\text{y}=55^\circ$
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