Solution

It is given that $A B | C D$ and $C D | E F$

$\therefore A B | C D|| E F$ (Lines parallel to the same line are parallel to each other) It can be observed that $x=z$

(Alternate interior angles) … (1)

It is given that $y: z=3: 7$

Let the common ratio between $y$ and $z$ be $a$. $\therefore$

$y=3 a$ and $z=7 a$

Also, $x+y=180^{\circ}$ (Co-interior angles on the same side of the transversal) $z$

$+y=180^{\circ}$ [Using equation (1)]

$7 a+3 a=180^{\circ}$

$10 a=180^{\circ} a=$

$18^{\circ} \therefore x=7 a=7 \times 18^{\circ}=$

Solution

It is given that, $A B | C D$

EF CD

$\therefore GED=126^{\circ}$

$\therefore \therefore$ GEF $+\therefore F E D=1260$

$\therefore \quad GEF+90^{\circ}=126^{\circ}$

$\therefore \therefore \quad GEF=36^{\circ}$

$\therefore \quad$ AGE and GED are alternate interior angles.

$\therefore \quad \therefore \quad$ AGE $=$ GED $=126^{\circ}$

Howềver, $AGÉ+FGE=180^{\circ}$ (Linear pair)

$\therefore \quad 1260^{\circ}+FGE=180^{\circ}$

$\therefore \quad \therefore \quad FGE=180^{\circ}-126^{\circ}=54^{\circ}$

$AGE=126^{\circ}, \therefore GEF=36^{\circ}, \therefore FGE=54^{\circ}$

Solution

Let us draw a line XY parallel to ST and passing through point $R$. $\therefore PQR+QRX=180^{\circ}$ (Co-interior angles on the same side of transversal QR)

$\therefore 110^{\circ}+Q R X=180^{\circ}$

$\therefore QRX=70^{\circ}$

Also,

$\therefore RST+\therefore$ SRY $=180^{\circ}$ (Co-interior angles on the same side of transversal SR) $0+S R Y=180^{\circ} 130$

$\therefore SRY=50^{\circ}$

$XY$ is a straight line. RQ and RS stand on it.

$\therefore \therefore QRX+\therefore QRS+\therefore SRY=180^{\circ} \circ+QRS+50^{\circ}=180^{\circ} 70$

$QRS=180^{\circ}-120^{\circ}=60^{\circ}$

$\therefore$

Solution

$\therefore APR=\therefore PRD$ (Alternate interior angles) In the given figure, if $AB | CD, APQ=50^{\circ}$ and $\therefore$ $P R D=1270$, find $x$ and $y$.

$50^{\circ}+y=127^{\circ} y=$

$127^{\circ}-50^{\circ} y=$

770

Also, $A P Q=P Q R$ (Alternate interior angles)

$50^{\circ}=x \quad x=50^{\circ}$ and $y=770$

Solution

Let us draw $BM \therefore PQ$ and $CN \therefore RS$.

As PQ || RS,

Therefore, BM || CN

Thus, $BM$ and $CN$ are two parallel lines and a transversal line $BC$ cuts them at $B$ and

C respectively.

$\therefore A=3$ (Alternate interior angles) 2

However, $1=2$ and $3=* 4$ (By"laws of reflection)

$\therefore 1=2 \doteq 3=4$

Also, $\dot{1}+2 \stackrel{\dot{\circ}}{=} 3+\dot{4}$

$\therefore ABC=\ddot{DCB}$

However, these are alternate interior angles. :

$A B|| C D$