Class 8: Lines and Angles

$\displaystyle \textbf{Question 1: }\text{In the adjoining figure, }AB\parallel CD.$
$\displaystyle \text{Find the values of }x\text{ and }y.\text{ Give reasons.}$
$\displaystyle \text{Answer:}$
$\displaystyle \angle EGC=\angle DGH \qquad\text{[Vertically opposite angles]}$
$\displaystyle \therefore \angle DGH=2x.$
$\displaystyle \angle DGH+\angle GHB=180^\circ \qquad\text{[Co-interior angles]}$
$\displaystyle 2x+3x=180$
$\displaystyle 5x=180$
$\displaystyle \therefore x=36^\circ.$

$\displaystyle \angle CGH=\angle GHB \qquad\text{[Alternate interior angles]}$
$\displaystyle \therefore y=3x$
$\displaystyle =3(36)$
$\displaystyle =108^\circ.$
$\displaystyle \therefore x=36^\circ\text{ and }y=108^\circ.$
$\displaystyle \text{Check: }2x+y=72^\circ+108^\circ=180^\circ.$

$\displaystyle \\$

$\displaystyle \textbf{Question 2: }\text{In each of the following figures, }AB\parallel CD.$
$\displaystyle \text{Find the value of }x.\text{ Give reasons.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) Since }AB\parallel CD,\text{ the co-interior angles are supplementary.}$
$\displaystyle \angle DGH+\angle BHG=180^\circ.$
$\displaystyle (4x-23)+3x=180$
$\displaystyle 7x-23=180$
$\displaystyle 7x=203$
$\displaystyle \therefore x=29^\circ.$
$\displaystyle \text{Check: }(4\times29-23)^\circ+(3\times29)^\circ$
$\displaystyle =93^\circ+87^\circ=180^\circ.$

$\displaystyle \text{(ii) Since }AB\parallel CD,\text{ corresponding angles are equal.}$
$\displaystyle \angle EDG=\angle GHB.$
$\displaystyle 2x+15=3x-20$
$\displaystyle x=35^\circ.$
$\displaystyle \text{Check: }(2\times35+15)^\circ=(3\times35-20)^\circ=85^\circ.$

$\displaystyle \\$

$\displaystyle \textbf{Question 3: }\text{In the adjoining figure, }AB\parallel CD$
$\displaystyle \text{are cut by a transversal at }E\text{ and }F\text{ respectively. If }$
$\displaystyle \angle2:\angle1=5:4,\text{ find the measure of each marked angle.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, }AB\parallel CD\text{ and }\angle2:\angle1=5:4.$
$\displaystyle \text{Let }\angle2=5x\text{ and }\angle1=4x.$
$\displaystyle \angle2+\angle1=180^\circ \qquad\text{[Linear pair]}$
$\displaystyle 5x+4x=180$
$\displaystyle 9x=180$
$\displaystyle \therefore x=20.$
$\displaystyle \therefore \angle2=5(20)^\circ=100^\circ$
$\displaystyle \text{and }\angle1=4(20)^\circ=80^\circ.$

$\displaystyle \angle3=\angle1=80^\circ \qquad\text{[Vertically opposite angles]}$
$\displaystyle \angle4=\angle2=100^\circ \qquad\text{[Vertically opposite angles]}$

$\displaystyle \angle6=\angle4=100^\circ \qquad\text{[Alternate interior angles]}$
$\displaystyle \angle5=\angle3=80^\circ \qquad\text{[Alternate interior angles]}$

$\displaystyle \angle7=\angle5=80^\circ \qquad\text{[Vertically opposite angles]}$
$\displaystyle \angle8=\angle6=100^\circ \qquad\text{[Vertically opposite angles]}$

$\displaystyle \therefore \angle1=\angle3=\angle5=\angle7=80^\circ,$
$\displaystyle \angle2=\angle4=\angle6=\angle8=100^\circ.$

$\displaystyle \\$

$\displaystyle \textbf{Question 4: }\text{In the adjoining figure, }AB\parallel CD.$
$\displaystyle \text{Find the values of }x,\ y\text{ and }z.$
$\displaystyle \text{Answer:}$
$\displaystyle \angle DPR=\angle PRQ \qquad\text{[Alternate interior angles]}$
$\displaystyle \therefore \angle PRQ=3x.$
$\displaystyle \text{In }\triangle PQR,$
$\displaystyle \angle PQR+\angle QPR+\angle PRQ=180^\circ \qquad\text{[Angle sum property]}$
$\displaystyle 80+2x+3x=180$
$\displaystyle 5x=100$
$\displaystyle \therefore x=20^\circ.$

$\displaystyle \therefore \angle QPR=2x=2(20)^\circ=40^\circ.$
$\displaystyle \angle PRQ=3x=3(20)^\circ=60^\circ.$
$\displaystyle \therefore y=60^\circ.$

$\displaystyle y+z=180^\circ \qquad\text{[Linear pair]}$
$\displaystyle 60+z=180$
$\displaystyle \therefore z=120^\circ.$
$\displaystyle \therefore x=20^\circ,\quad y=60^\circ,\quad z=120^\circ.$
$\displaystyle \text{Check: }80^\circ+40^\circ+60^\circ=180^\circ.$

$\displaystyle \\$

$\displaystyle \textbf{Question 5: }\text{In each of the following figures, }AB\parallel CD.$
$\displaystyle \text{Find the value of }x\text{ in each case.}$
$\displaystyle \text{Answer:}$

$\displaystyle \text{(i) Draw a line through }E\text{ parallel to }AB\text{ and }CD.$
$\displaystyle \text{At }A,\text{ the acute angle made by }AE\text{ with the parallel line is}$
$\displaystyle 180^\circ-104^\circ=76^\circ.$
$\displaystyle \text{At }C,\text{ the acute angle made by }CE\text{ with the parallel line is}$
$\displaystyle 180^\circ-116^\circ=64^\circ.$
$\displaystyle \therefore x=76^\circ+64^\circ$
$\displaystyle =140^\circ.$
$\displaystyle \therefore x=140^\circ.$

$\displaystyle \text{(ii) Draw a line through }E\text{ parallel to }AB\text{ and }CD.$
$\displaystyle \text{The upper part of }x=35^\circ \qquad\text{[Alternate interior angles]}$
$\displaystyle \text{The lower part of }x=65^\circ \qquad\text{[Alternate interior angles]}$
$\displaystyle \therefore x=35^\circ+65^\circ=100^\circ.$

$\displaystyle \text{(iii) Draw a line through }M\text{ parallel to }AB\text{ and }CD.$
$\displaystyle \text{The angle made by }BM\text{ with this parallel line is }35^\circ.$
$\displaystyle \text{The angle made by }DM\text{ with this parallel line is }75^\circ.$
$\displaystyle \text{Therefore, the smaller angle between }BM\text{ and }DM\text{ is}$
$\displaystyle 35^\circ+75^\circ=110^\circ.$
$\displaystyle \text{Since }x\text{ is the reflex angle at }M,$
$\displaystyle x=360^\circ-110^\circ$
$\displaystyle =250^\circ.$
$\displaystyle \therefore x=250^\circ.$

$\displaystyle \text{(iv) Draw a line through the lower vertex parallel to }AB\text{ and }CD.$
$\displaystyle \text{The angle on the left is }63^\circ \qquad\text{[Corresponding angles]}$
$\displaystyle \text{The acute angle on the right is }180^\circ-130^\circ=50^\circ.$
$\displaystyle \text{Therefore, the smaller angle between the two slanting lines is}$
$\displaystyle 63^\circ+50^\circ=113^\circ.$
$\displaystyle \text{Since }x\text{ is the reflex angle,}$
$\displaystyle x=360^\circ-113^\circ$
$\displaystyle =247^\circ.$
$\displaystyle \therefore x=247^\circ.$

$\displaystyle \text{(v) Since }AB\parallel CD,\text{ corresponding angles are equal.}$ q10
$\displaystyle \text{The angle corresponding to }3(x-6)^\circ\text{ is also }3(x-6)^\circ.$
$\displaystyle \text{At the lower intersection, the vertically opposite angle is}$
$\displaystyle (x+5)^\circ+27^\circ=(x+32)^\circ.$
$\displaystyle \therefore 3(x-6)=x+32$
$\displaystyle 3x-18=x+32$
$\displaystyle 2x=50$
$\displaystyle \therefore x=25^\circ.$
$\displaystyle \text{Check: }3(25-6)^\circ=57^\circ\text{ and }25^\circ+5^\circ+27^\circ=57^\circ.$

$\displaystyle \text{(vi) Since }AB\parallel CD,\text{ the co-interior angles are supplementary.}$ q11
$\displaystyle \text{The upper interior angle is }(2x-10)^\circ+32^\circ.$
$\displaystyle \text{The lower interior angle is }(3x+38)^\circ.$
$\displaystyle (2x-10)+32+(3x+38)=180$
$\displaystyle 5x+60=180$
$\displaystyle 5x=120$
$\displaystyle \therefore x=24^\circ.$

$\displaystyle \\$

$\displaystyle \textbf{Question 6: }\text{In the adjoining figure, }AB\parallel CD.$ q13
$\displaystyle \text{Find the values of }x,\ y,\ z.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Since }A,\ E,\ B\text{ lie on a straight line,}$
$\displaystyle x+70+x=180^\circ.$
$\displaystyle 2x+70=180$
$\displaystyle 2x=110$
$\displaystyle \therefore x=55^\circ.$

$\displaystyle \text{Since }AB\parallel CD,\text{ and }CE\text{ is a transversal,}$
$\displaystyle x+z=180^\circ \qquad\text{[Co-interior angles]}$
$\displaystyle 55+z=180$
$\displaystyle \therefore z=125^\circ.$

$\displaystyle \text{In }\triangle APE,\ \angle PAE=90^\circ.$
$\displaystyle \angle PAE+\angle AEP+\angle APE=180^\circ \qquad\text{[Angle sum property]}$
$\displaystyle 90+55+y=180$
$\displaystyle \therefore y=35^\circ.$

$\displaystyle \therefore x=55^\circ,\quad y=35^\circ,\quad z=125^\circ.$

$\displaystyle \\$

$\displaystyle \textbf{Question 7: }\text{In each of the following figures, }AB\parallel CD.$
$\displaystyle \text{Find the values of }x,\ y,\ z.$
$\displaystyle \text{Answer:}$

$\displaystyle \text{(i) Since }AB\parallel CD,$ q14
$\displaystyle \angle BAO=\angle ODC \qquad\text{[Alternate interior angles]}$
$\displaystyle \therefore x=40^\circ.$
$\displaystyle \angle ABO=\angle OCD \qquad\text{[Alternate interior angles]}$
$\displaystyle \therefore z=80^\circ.$
$\displaystyle \text{In }\triangle CDO,$
$\displaystyle x+80+y=180^\circ \qquad\text{[Angle sum property]}$
$\displaystyle 40+80+y=180$
$\displaystyle \therefore y=60^\circ.$
$\displaystyle \therefore x=40^\circ,\quad y=60^\circ,\quad z=80^\circ.$

$\displaystyle \text{(ii) Since }AB\parallel CD,$ q15
$\displaystyle \angle CGH+\angle AHG=180^\circ \qquad\text{[Co-interior angles]}$
$\displaystyle 70+x=180$
$\displaystyle \therefore x=110^\circ.$
$\displaystyle \text{Since }x\text{ and }y\text{ form a linear pair,}$
$\displaystyle x+y=180^\circ.$
$\displaystyle 110+y=180$
$\displaystyle \therefore y=70^\circ.$
$\displaystyle \text{At }J,\text{ the angle adjacent to }120^\circ\text{ is}$
$\displaystyle 180^\circ-120^\circ=60^\circ.$
$\displaystyle \text{In }\triangle GHJ,$
$\displaystyle z+y+60^\circ=180^\circ \qquad\text{[Angle sum property]}$
$\displaystyle z+70+60=180$
$\displaystyle \therefore z=50^\circ.$
$\displaystyle \therefore x=110^\circ,\quad y=70^\circ,\quad z=50^\circ.$

$\displaystyle \\$

$\displaystyle \textbf{Question 8: }\text{In the given figure, }AB\parallel CD\text{ and }$ q16
$\displaystyle EF\parallel GH.\text{ Find the values of }x,\ y,\ z,\ t\text{ and }w.$
$\displaystyle \text{Answer:}$
$\displaystyle x=60^\circ \qquad\text{[Vertically opposite angles]}$
$\displaystyle x=y \qquad\text{[Alternate interior angles, since }EF\parallel GH\text{]}$
$\displaystyle \therefore y=60^\circ.$

$\displaystyle \text{The angle supplementary to }110^\circ\text{ is}$
$\displaystyle 180^\circ-110^\circ=70^\circ.$
$\displaystyle \text{In the triangle formed above }AB,$
$\displaystyle 70^\circ+60^\circ+\angle CYZ=180^\circ \qquad\text{[Angle sum property]}$
$\displaystyle \angle CYZ=180^\circ-70^\circ-60^\circ=50^\circ.$

$\displaystyle \text{At }Y,\text{ the angles }50^\circ,\ y^\circ\text{ and }z^\circ \text{ form a straight angle.}$
$\displaystyle 50+60+z=180$
$\displaystyle \therefore z=70^\circ.$

$\displaystyle \text{Since }AB\parallel CD,\text{ and }GH\text{ is a transversal,}$
$\displaystyle z+t=180^\circ \qquad\text{[Co-interior angles]}$
$\displaystyle 70+t=180$
$\displaystyle \therefore t=110^\circ.$

$\displaystyle t+w=180^\circ \qquad\text{[Linear pair]}$
$\displaystyle 110+w=180$
$\displaystyle \therefore w=70^\circ.$

$\displaystyle \therefore x=60^\circ,\quad y=60^\circ,\quad z=70^\circ,$
$\displaystyle t=110^\circ,\quad w=70^\circ.$

$\displaystyle \\$

$\displaystyle \textbf{Question 9: }\text{In the given figure, }AB\parallel CD,\ EF\parallel GH$ q17
$\displaystyle \text{and }JK\parallel LM.\text{ Find the values of }x,\ y,\ z,\ t \text{ and }w.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Since }AB\parallel CD,\text{ and }LM\text{ is a transversal,}$
$\displaystyle w=70^\circ \qquad\text{[Alternate interior angles]}$

$\displaystyle \text{Since }EF\parallel GH,\text{ and }LM\text{ is a transversal,}$
$\displaystyle x=45^\circ \qquad\text{[Alternate interior angles]}$

$\displaystyle \text{At }R,\text{ the angles }y,\ x\text{ and }70^\circ \text{ form a straight angle.}$
$\displaystyle y+x+70=180$
$\displaystyle y+45+70=180$
$\displaystyle \therefore y=65^\circ.$

$\displaystyle \text{Since }AB\parallel CD,\text{ and }EF\text{ is a transversal,}$
$\displaystyle t=y \qquad\text{[Alternate interior angles]}$
$\displaystyle \therefore t=65^\circ.$

$\displaystyle \text{Since }JK\parallel LM,\text{ and }EF\text{ is a transversal,}$
$\displaystyle z=x \qquad\text{[Alternate interior angles]}$
$\displaystyle \therefore z=45^\circ.$

$\displaystyle \therefore x=45^\circ,\quad y=65^\circ,\quad z=45^\circ,$
$\displaystyle t=65^\circ,\quad w=70^\circ.$

$\displaystyle \\$

$\displaystyle \textbf{Question 10: }\text{In each of the following figures, find the value}$
$\displaystyle \text{of }x\text{ for which the lines }l\text{ and }m\text{ will be parallel.}$
$\displaystyle \text{Answer:}$

$\displaystyle \text{(i) For }l\text{ and }m\text{ to be parallel,}$ q18
$\displaystyle (3x-25)^\circ=(2x+5)^\circ \qquad\text{[Corresponding angles]}$
$\displaystyle 3x-25=2x+5$
$\displaystyle \therefore x=30^\circ.$

$\displaystyle \text{(ii) For }l\text{ and }m\text{ to be parallel,}$ q19
$\displaystyle (2x-10)^\circ=(5x-20)^\circ \qquad\text{[Alternate exterior angles]}$
$\displaystyle 2x-10=5x-20$
$\displaystyle 3x=10$
$\displaystyle \therefore x=\frac{10}{3}^\circ.$

$\displaystyle \text{(iii) For }l\text{ and }m\text{ to be parallel,}$ q20
$\displaystyle \text{the angle corresponding to }70^\circ\text{ must also be }70^\circ.$
$\displaystyle (3x+14)^\circ+(2x+6)^\circ=70^\circ.$
$\displaystyle 3x+14+2x+6=70$
$\displaystyle 5x+20=70$
$\displaystyle 5x=50$
$\displaystyle \therefore x=10^\circ.$

$\displaystyle \text{(iv) From the angles on the straight line }l,$ q21
$\displaystyle 4y+3y+5y=180$
$\displaystyle 12y=180$
$\displaystyle \therefore y=15^\circ.$
$\displaystyle \text{For }l\text{ and }m\text{ to be parallel,}$
$\displaystyle x=(3y+5y)^\circ \qquad\text{[Alternate interior angles]}$
$\displaystyle x=8y$
$\displaystyle =8(15)$
$\displaystyle \therefore x=120^\circ.$

$\displaystyle \\$

$\displaystyle \textbf{Question 11: }\text{In the adjoining figure, }AB\parallel CD$ q22
$\displaystyle \text{and they cut the lines }PQ\text{ and }QR\text{ at }E,\ G \text{ and }F,\ H$
$\displaystyle \text{respectively. Find the value of }x.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Since }AB\parallel CD,\text{ and }PQ\text{ is a transversal,}$
$\displaystyle \angle PEF=\angle DGE=85^\circ \qquad\text{[Corresponding angles]}$
$\displaystyle \angle DGE+\angle DGH=180^\circ \qquad\text{[Linear pair]}$
$\displaystyle \therefore \angle DGH=180^\circ-85^\circ=95^\circ.$

$\displaystyle \angle QHG=180^\circ-125^\circ=55^\circ \qquad\text{[Linear pair]}$

$\displaystyle \text{In }\triangle GQH,$
$\displaystyle \angle HGQ+\angle QHG+\angle GQH=180^\circ \qquad\text{[Angle sum property]}$
$\displaystyle 95+55+x=180$
$\displaystyle x=180-150$
$\displaystyle \therefore x=30^\circ.$

$\displaystyle \\$

$\displaystyle \textbf{Question 12: }\text{In the adjoining figure, }AB\parallel EF$ q23
$\displaystyle \text{and }EF\parallel CD.\text{ Find the value of }x.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, }AB\parallel EF\parallel CD.$
$\displaystyle \angle BCD=\angle ABC=70^\circ \qquad\text{[Alternate interior angles]}$
$\displaystyle \angle ECD=180^\circ-130^\circ=50^\circ \qquad\text{[Co-interior angles]}$
$\displaystyle \angle BCE+\angle ECD=\angle BCD.$
$\displaystyle x+50=70$
$\displaystyle \therefore x=20^\circ.$

$\displaystyle \\$

$\displaystyle \textbf{Question 13: }\text{In the adjoining figure, }AB\parallel CD.$ q24
$\displaystyle \text{Find the value of }x.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, }AB\parallel CD.$
$\displaystyle \text{Draw a line through }G\text{ parallel to }AB\text{ and }CD.$

$\displaystyle \text{The angle between }EG\text{ and the parallel through }G\text{ is }20^\circ$
$\displaystyle \qquad\text{[Alternate interior angles]}.$

$\displaystyle \angle BEF=180^\circ-75^\circ=105^\circ \qquad\text{[Linear pair]}$
$\displaystyle \angle EFD=180^\circ-105^\circ=75^\circ \qquad\text{[Co-interior angles]}$

$\displaystyle \angle GFD=\angle EFD-\angle EFG$
$\displaystyle =75^\circ-25^\circ$
$\displaystyle =50^\circ.$

$\displaystyle \text{The angle between }FG\text{ and the parallel through }G\text{ is }50^\circ$
$\displaystyle \qquad\text{[Alternate interior angles]}.$

$\displaystyle \therefore x=20^\circ+50^\circ$
$\displaystyle =70^\circ.$
$\displaystyle \therefore \text{The required value is }x=70^\circ.$

$\displaystyle \\$

$\displaystyle \textbf{Question 14: }\text{In the adjoining figure, }AB\parallel CD \text{ and }EF$ q25
$\displaystyle \text{cuts them at }G\text{ and }H\text{ respectively. }GP\text{ and }HQ$
$\displaystyle \text{are the bisectors of }\angle AGH\text{ and }\angle GHD \text{ respectively.}$
$\displaystyle \text{Prove that }GP\parallel HQ.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, }AB\parallel CD.$
$\displaystyle GP\text{ bisects }\angle AGH.$
$\displaystyle \therefore \angle PGH=\frac{1}{2}\angle AGH.$
$\displaystyle HQ\text{ bisects }\angle GHD.$
$\displaystyle \therefore \angle QHG=\frac{1}{2}\angle GHD.$
$\displaystyle \text{Since }AB\parallel CD\text{ and }EF\text{ is a transversal,}$
$\displaystyle \angle AGH=\angle GHD \qquad\text{[Alternate interior angles]}$
$\displaystyle \therefore \frac{1}{2}\angle AGH=\frac{1}{2}\angle GHD.$
$\displaystyle \therefore \angle PGH=\angle QHG.$
$\displaystyle \text{These are alternate interior angles made by }GP\text{ and }HQ$
$\displaystyle \text{with the transversal }GH.$
$\displaystyle \therefore GP\parallel HQ \qquad\text{[Converse of alternate interior angle theorem]}.$
$\displaystyle \text{Hence proved.}$

$\displaystyle \\$

$\displaystyle \textbf{Question 15: }\text{In each of the following figures, determine}$
$\displaystyle \text{the values of }x,\ y,\ z\text{ and }t.$
$\displaystyle \text{Answer:}$

$\displaystyle \text{(i)}$ q27
$\displaystyle t=67^\circ \qquad\text{[Corresponding angles]}$
$\displaystyle x+67^\circ=180^\circ \qquad\text{[Co-interior angles]}$
$\displaystyle \therefore x=113^\circ.$
$\displaystyle x+z=180^\circ \qquad\text{[Linear pair]}$
$\displaystyle 113+z=180$
$\displaystyle \therefore z=67^\circ.$
$\displaystyle x+y=180^\circ \qquad\text{[Linear pair]}$
$\displaystyle 113+y=180$
$\displaystyle \therefore y=67^\circ.$
$\displaystyle \therefore x=113^\circ,\quad y=67^\circ,\quad z=67^\circ,\quad t=67^\circ.$

$\displaystyle \text{(ii)}$ q28
$\displaystyle 2x+110^\circ=180^\circ \qquad\text{[Linear pair]}$
$\displaystyle 2x=70$
$\displaystyle \therefore x=35^\circ.$
$\displaystyle 5t=2x \qquad\text{[Corresponding angles]}$
$\displaystyle 5t=70$
$\displaystyle \therefore t=14^\circ.$
$\displaystyle (3z-25)^\circ+5t^\circ=180^\circ \qquad\text{[Co-interior angles]}$
$\displaystyle 3z-25+70=180$
$\displaystyle 3z=135$
$\displaystyle \therefore z=45^\circ.$
$\displaystyle (3z-25)+(2y-10)=180$
$\displaystyle 110+2y-10=180$
$\displaystyle 2y=80$
$\displaystyle \therefore y=40^\circ.$
$\displaystyle \therefore x=35^\circ,\quad y=40^\circ,\quad z=45^\circ,\quad t=14^\circ.$

$\displaystyle \text{(iii)}$ q29
$\displaystyle y+72^\circ=180^\circ \qquad\text{[Co-interior angles]}$
$\displaystyle \therefore y=108^\circ.$
$\displaystyle 72^\circ+3z^\circ=180^\circ \qquad\text{[Co-interior angles]}$
$\displaystyle 3z=108$
$\displaystyle \therefore z=36^\circ.$
$\displaystyle t+108^\circ=180^\circ \qquad\text{[Linear pair]}$
$\displaystyle \therefore t=72^\circ.$
$\displaystyle 132^\circ+x=180^\circ \qquad\text{[Linear pair]}$
$\displaystyle \therefore x=48^\circ.$
$\displaystyle \therefore x=48^\circ,\quad y=108^\circ,\quad z=36^\circ,\quad t=72^\circ.$

$\displaystyle \text{(iv)}$ q30
$\displaystyle (2y-11)^\circ+5z^\circ=180^\circ \qquad\text{[Co-interior angles]}$
$\displaystyle 2y+5z=191 \qquad\ldots\text{(i)}$
$\displaystyle 5z^\circ=180^\circ-55^\circ \qquad\text{[Co-interior angles]}$
$\displaystyle 5z=125$
$\displaystyle \therefore z=25^\circ.$
$\displaystyle (4t+7)^\circ=125^\circ \qquad\text{[Corresponding angles]}$
$\displaystyle 4t=118$
$\displaystyle \therefore t=\frac{59}{2}^\circ=29.5^\circ.$
$\displaystyle (x+25)^\circ+55^\circ=180^\circ \qquad\text{[Co-interior angles]}$
$\displaystyle x+80=180$
$\displaystyle \therefore x=100^\circ.$
$\displaystyle \text{Substituting }z=25\text{ in (i),}$
$\displaystyle 2y+5(25)=191$
$\displaystyle 2y=66$
$\displaystyle \therefore y=33^\circ.$
$\displaystyle \therefore x=100^\circ,\quad y=33^\circ,\quad z=25^\circ,\quad t=29.5^\circ.$

$\displaystyle \\$

$\displaystyle \textbf{Question 16: }\text{State, giving reasons, whether }AB\parallel CD \text{ or not.}$
$\displaystyle \text{Given in (iii): }EF\parallel GH.$
$\displaystyle \text{Answer:}$

$\displaystyle \text{(i)}$ q27
$\displaystyle \angle EGD=67^\circ \qquad\text{[Vertically opposite angles]}$
$\displaystyle \angle GHB=67^\circ \qquad\text{[Corresponding angles]}$
$\displaystyle \therefore \angle EGD=\angle GHB.$
$\displaystyle \text{Hence }AB\parallel CD \qquad\text{[Converse of alternate interior angle theorem]}.$

$\displaystyle \text{(ii)}$ q32
$\displaystyle \angle AHG=100^\circ,\qquad \angle HGC=70^\circ.$
$\displaystyle \angle AHG+\angle HGC=100^\circ+70^\circ=170^\circ.$
$\displaystyle 170^\circ\ne180^\circ.$
$\displaystyle \therefore AB\text{ is not parallel to }CD \qquad\text{[Co-interior angles are not supplementary]}.$

$\displaystyle \text{(iii)}$ q34
$\displaystyle \text{Given, }EF\parallel GH.$
$\displaystyle \angle AEF=100^\circ.$
$\displaystyle \therefore \angle BHG=100^\circ \qquad\text{[Corresponding angles]}$
$\displaystyle \angle GHC=100^\circ.$
$\displaystyle \angle GHC+\angle CHF=100^\circ+100^\circ=200^\circ.$
$\displaystyle 200^\circ\ne180^\circ.$
$\displaystyle \therefore AB\text{ is not parallel to }CD \qquad\text{[Co-interior angles are not supplementary]}.$

$\displaystyle \text{(iv)}$ q35
$\displaystyle \angle AYE=180^\circ-64^\circ=116^\circ \qquad\text{[Linear pair]}.$
$\displaystyle \angle CYX=180^\circ-126^\circ=54^\circ \qquad\text{[Linear pair]}.$
$\displaystyle \angle AYE+\angle CYX=116^\circ+54^\circ=170^\circ.$
$\displaystyle 170^\circ\ne180^\circ.$
$\displaystyle \therefore AB\text{ is not parallel to }CD \qquad\text{[Co-interior angles are not supplementary]}.$

$\displaystyle \\$

$\displaystyle \textbf{Question 17: }\text{In the given figure, }AB\parallel CD \text{ and }\angle A=\angle C.$ q36
$\displaystyle \text{Prove that }AE\parallel CF.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, }AB\parallel CD\text{ and }\angle A=\angle C.$
$\displaystyle \text{Since }AB\parallel CD\text{ and }AE\text{ is a transversal,}$
$\displaystyle \angle BAE=\angle CDE \qquad\text{[Corresponding angles]}.$
$\displaystyle \text{But }\angle BAE=\angle DCF \qquad\text{[Given]}.$
$\displaystyle \therefore \angle CDE=\angle DCF.$
$\displaystyle \text{Hence }AE\parallel CF \qquad\text{[Converse of corresponding angle theorem]}.$
$\displaystyle \text{Hence proved.}$