In the figure,if $AB \parallel CD$,$EF \perp CD$ and $\angle GED = 126^o$,find $\angle AGE$,$\angle GEF$ and $\angle FGE$.
(N/A) Given that $AB \parallel CD$,$EF \perp CD$,and $\angle GED = 126^o$.
$1$. Since $AB \parallel CD$ and $GE$ is a transversal,the alternate interior angles are equal.
Therefore,$\angle AGE = \angle GED$.
Given $\angle GED = 126^o$,so $\angle AGE = 126^o$.
$2$. We know that $\angle GED = \angle GEF + \angle FED$.
Since $EF \perp CD$,$\angle FED = 90^o$.
Therefore,$126^o = \angle GEF + 90^o$.
$\angle GEF = 126^o - 90^o = 36^o$.
$3$. Since $AB \parallel CD$ and $GE$ is a transversal,the sum of interior angles on the same side of the transversal is $180^o$.
Therefore,$\angle FGE + \angle GED = 180^o$.
$\angle FGE + 126^o = 180^o$.
$\angle FGE = 180^o - 126^o = 54^o$.
Thus,$\angle AGE = 126^o$,$\angle GEF = 36^o$,and $\angle FGE = 54^o$.
$1$. Since $AB \parallel CD$ and $GE$ is a transversal,the alternate interior angles are equal.
Therefore,$\angle AGE = \angle GED$.
Given $\angle GED = 126^o$,so $\angle AGE = 126^o$.
$2$. We know that $\angle GED = \angle GEF + \angle FED$.
Since $EF \perp CD$,$\angle FED = 90^o$.
Therefore,$126^o = \angle GEF + 90^o$.
$\angle GEF = 126^o - 90^o = 36^o$.
$3$. Since $AB \parallel CD$ and $GE$ is a transversal,the sum of interior angles on the same side of the transversal is $180^o$.
Therefore,$\angle FGE + \angle GED = 180^o$.
$\angle FGE + 126^o = 180^o$.
$\angle FGE = 180^o - 126^o = 54^o$.
Thus,$\angle AGE = 126^o$,$\angle GEF = 36^o$,and $\angle FGE = 54^o$.
Explore More
Similar Questions
In the figure,$\angle X = 62^o$ and $\angle XYZ = 54^o$. If $YO$ and $ZO$ are the bisectors of $\angle XYZ$ and $\angle XZY$ respectively of $\Delta XYZ$,find $\angle OZY$ and $\angle YOZ$.
Difficult
View SolutionIn the figure,ray $OS$ stands on a line $POQ$. Ray $OR$ and ray $OT$ are angle bisectors of $\angle POS$ and $\angle SOQ$,respectively. If $\angle POS = x$,find $\angle ROT$. (in $^o$)
Medium
View SolutionIf a transversal intersects two lines such that the bisectors of a pair of corresponding angles are parallel,then prove that the two lines are parallel.
Medium
View SolutionIn the figure,sides $QP$ and $RQ$ of $\Delta PQR$ are produced to points $S$ and $T$ respectively. If $\angle SPR = 135^o$ and $\angle PQT = 110^o$,find $\angle PRQ$. (in $^o$)
Medium
View SolutionIn the figure,if $AB \parallel CD$,$CD \parallel EF$ and $y: z = 3: 7$,find $x$. (in $^o$)
Difficult
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo