In the figure,if $AB \parallel CD$,$EF \perp CD$ and $\angle GED = 126^o$,find $\angle AGE$,$\angle GEF$ and $\angle FGE$.

Vedclass pdf generator app on play store
Vedclass iOS app on app store
(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 $\angle FGE = 54^o$.

Explore More

Similar Questions

In the figure,the side $QR$ of $\Delta PQR$ is produced to a point $S$. If the bisectors of $\angle PQR$ and $\angle PRS$ meet at point $T$,then prove that $\angle QTR = \frac{1}{2} \angle QPR$.

In the figure,if lines $PQ$ and $RS$ intersect at point $T$,such that $\angle PRT = 40^\circ$,$\angle RPT = 95^\circ$ and $\angle TSQ = 75^\circ$,find $\angle SQT$. (in $^\circ$)

In the figure,if $PQ \parallel ST$,$\angle PQR = 110^o$ and $\angle RST = 130^o$,find $\angle QRS$. (in $^o$)

Difficult
View Solution

In the figure,find the values of $x$ and $y$ and then show that $AB \parallel CD$.

In the figure,if $PQ \perp PS$,$PQ \parallel SR$,$\angle SQR = 28^o$ and $\angle QRT = 65^o$,then find the values of $x$ and $y$.

Difficult
View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo