Question
In the given figure, $AB \| CD \| EF , \angle D B G=x, \angle E D H=y, \angle A E B=z, \angle E A B=90^{\circ}$ and $\angle B E F=65^{\circ}$. Find the values of $x , y$ and z .
✓
Answer
$EF \| CD$ and $ED$ is the transversal.
$\therefore \angle F E D+\angle E D H=180^{\circ}$ [co-interior angles]
$\Rightarrow 65^{\circ}+y=180^{\circ}$
$\Rightarrow y=\left(180^{\circ}-65^{\circ}\right)=115^{\circ}$
Now $C H \| A G$ and $D B$ is the transversal
$\therefore x = y =115^{\circ} [$corresponding angles$]$
Now, ABG is a straight line.
$\therefore \angle A B E+\angle E B G=180^{\circ}[$ sum of linear pair of angles is $180^{\circ}]$
$\Rightarrow \angle A B E+x=180^{\circ}$
$\Rightarrow \angle A B E+115^{\circ}=180^{\circ}$
$\Rightarrow \angle A B E=\left(180^{\circ}-115^{\circ}\right)=65^{\circ}$
We know that the sum of the angles of a triangle is $180^{\circ}$.
From $\triangle E A B$, we get
$\angle E A B+\angle A B E+\angle B E A=180^{\circ}$
$\Rightarrow 90^{\circ}+65^{\circ}+ z =180^{\circ}$
$\Rightarrow z =\left(180^{\circ}-155^{\circ}\right)=25^{\circ}$
$\therefore x =115^{\circ}, y =115^{\circ}$ and $z =25^{\circ}$
$\therefore \angle F E D+\angle E D H=180^{\circ}$ [co-interior angles]
$\Rightarrow 65^{\circ}+y=180^{\circ}$
$\Rightarrow y=\left(180^{\circ}-65^{\circ}\right)=115^{\circ}$
Now $C H \| A G$ and $D B$ is the transversal
$\therefore x = y =115^{\circ} [$corresponding angles$]$
Now, ABG is a straight line.
$\therefore \angle A B E+\angle E B G=180^{\circ}[$ sum of linear pair of angles is $180^{\circ}]$
$\Rightarrow \angle A B E+x=180^{\circ}$
$\Rightarrow \angle A B E+115^{\circ}=180^{\circ}$
$\Rightarrow \angle A B E=\left(180^{\circ}-115^{\circ}\right)=65^{\circ}$
We know that the sum of the angles of a triangle is $180^{\circ}$.
From $\triangle E A B$, we get
$\angle E A B+\angle A B E+\angle B E A=180^{\circ}$
$\Rightarrow 90^{\circ}+65^{\circ}+ z =180^{\circ}$
$\Rightarrow z =\left(180^{\circ}-155^{\circ}\right)=25^{\circ}$
$\therefore x =115^{\circ}, y =115^{\circ}$ and $z =25^{\circ}$
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