Page 1
Question:29
In the given figure, l || m and a transversal t cuts them. If ?1 = 120°, find the measure of each of the remaining marked angles.
Solution:
We have, ?1 = 120°
. Then,
?1 = ?5 [Corresponding angles] ? ?5 = 120° ?1 = ?3 [Vertically-opposite angles] ? ?3 = 120°
?5 = ?7 [Vertically-opposite angles] ? ?7 = 120° ?1 + ?2 = 180° [Since AFB is a straight line] ? 120°+ ?2 = 180°
? ?2 = 60° ?2 = ?4 [Vertically-opposite angles] ? ?4 = 60° ?2 = ?6 [Corresponding angles]
? ?6 = 60° ?6 = ?8 [Vertically-opposite angles] ? ?8 = 60° ? ?1 = 120°, ?2 = 60°, ?3 = 120°, ?4 = 60°, ?5 = 120°, ?6 = 60°, ?7 = 120° and ?8 = 60°
Question:30
In the given figure, l || m and a transversal t cuts them. If ?7 = 80°, find the measure of each of the remaining marked angles.
Solution:
In the given figure, ?7 and ?8 form a linear pair.
? ?7 + ?8 = 180º
? 80º + ?8 = 180º
? ?8 = 180º - 80º = 100º
Now,
?6 = ?8 = 100º Verticallyoppositeangles
?5 = ?7 = 80º Verticallyoppositeangles
It is given that, l || m and t is a transversal.
? ?1 = ?5 = 80º Pairofcorrespondingangles
?2 = ?6 = 100º Pairofcorrespondingangles
?3 = ?7 = 80º Pairofcorrespondingangles
?4 = ?8 = 100º Pairofcorrespondingangles
Question:31
In the given figure, l || m and a transversal t cuts them. If ?1 : ?2 = 2 : 3, find the measure of each of the marked angles.
Page 2
Question:29
In the given figure, l || m and a transversal t cuts them. If ?1 = 120°, find the measure of each of the remaining marked angles.
Solution:
We have, ?1 = 120°
. Then,
?1 = ?5 [Corresponding angles] ? ?5 = 120° ?1 = ?3 [Vertically-opposite angles] ? ?3 = 120°
?5 = ?7 [Vertically-opposite angles] ? ?7 = 120° ?1 + ?2 = 180° [Since AFB is a straight line] ? 120°+ ?2 = 180°
? ?2 = 60° ?2 = ?4 [Vertically-opposite angles] ? ?4 = 60° ?2 = ?6 [Corresponding angles]
? ?6 = 60° ?6 = ?8 [Vertically-opposite angles] ? ?8 = 60° ? ?1 = 120°, ?2 = 60°, ?3 = 120°, ?4 = 60°, ?5 = 120°, ?6 = 60°, ?7 = 120° and ?8 = 60°
Question:30
In the given figure, l || m and a transversal t cuts them. If ?7 = 80°, find the measure of each of the remaining marked angles.
Solution:
In the given figure, ?7 and ?8 form a linear pair.
? ?7 + ?8 = 180º
? 80º + ?8 = 180º
? ?8 = 180º - 80º = 100º
Now,
?6 = ?8 = 100º Verticallyoppositeangles
?5 = ?7 = 80º Verticallyoppositeangles
It is given that, l || m and t is a transversal.
? ?1 = ?5 = 80º Pairofcorrespondingangles
?2 = ?6 = 100º Pairofcorrespondingangles
?3 = ?7 = 80º Pairofcorrespondingangles
?4 = ?8 = 100º Pairofcorrespondingangles
Question:31
In the given figure, l || m and a transversal t cuts them. If ?1 : ?2 = 2 : 3, find the measure of each of the marked angles.
Solution:
Let ?1 = 2k and ?2 = 3k, where k is some constant.
Now, ?1 and ?2 form a linear pair.
? ?1 + ?2 = 180º
? 2k + 3k = 180º
? 5k = 180º
? k = 36º
? ?1 = 2k = 2 × 36º = 72º
?2 = 3k = 3 × 36º = 108º
Now,
?3 = ?1 = 72º Verticallyoppositeangles
?4 = ?2 = 108º Verticallyoppositeangles
It is given that, l || m and t is a transversal.
? ?5 = ?1 = 72º Pairofcorrespondingangles
?6 = ?2 = 108º Pairofcorrespondingangles
?7 = ?1 = 72º Pairofalternateexteriorangles
?8 = ?2 = 108º Pairofalternateexteriorangles
Question:32
For what value of x will the line l and m be parallel to each other?
Solution:
For the lines l and m to be parallel
? 3x -20 = 2x +10 [Corresponding Angles] ? x = 30
Question:33
For what value of x will the lines l and m be parallel to each other?
Solution:
? 3x +5 +4x = 180 [Consecutive Interior Angles] ? 7x = 175 ? x = 25
Question:34
In the given figure, AB || CD and BC || ED. Find the value of x.
Solution:
BC ? ED
Page 3
Question:29
In the given figure, l || m and a transversal t cuts them. If ?1 = 120°, find the measure of each of the remaining marked angles.
Solution:
We have, ?1 = 120°
. Then,
?1 = ?5 [Corresponding angles] ? ?5 = 120° ?1 = ?3 [Vertically-opposite angles] ? ?3 = 120°
?5 = ?7 [Vertically-opposite angles] ? ?7 = 120° ?1 + ?2 = 180° [Since AFB is a straight line] ? 120°+ ?2 = 180°
? ?2 = 60° ?2 = ?4 [Vertically-opposite angles] ? ?4 = 60° ?2 = ?6 [Corresponding angles]
? ?6 = 60° ?6 = ?8 [Vertically-opposite angles] ? ?8 = 60° ? ?1 = 120°, ?2 = 60°, ?3 = 120°, ?4 = 60°, ?5 = 120°, ?6 = 60°, ?7 = 120° and ?8 = 60°
Question:30
In the given figure, l || m and a transversal t cuts them. If ?7 = 80°, find the measure of each of the remaining marked angles.
Solution:
In the given figure, ?7 and ?8 form a linear pair.
? ?7 + ?8 = 180º
? 80º + ?8 = 180º
? ?8 = 180º - 80º = 100º
Now,
?6 = ?8 = 100º Verticallyoppositeangles
?5 = ?7 = 80º Verticallyoppositeangles
It is given that, l || m and t is a transversal.
? ?1 = ?5 = 80º Pairofcorrespondingangles
?2 = ?6 = 100º Pairofcorrespondingangles
?3 = ?7 = 80º Pairofcorrespondingangles
?4 = ?8 = 100º Pairofcorrespondingangles
Question:31
In the given figure, l || m and a transversal t cuts them. If ?1 : ?2 = 2 : 3, find the measure of each of the marked angles.
Solution:
Let ?1 = 2k and ?2 = 3k, where k is some constant.
Now, ?1 and ?2 form a linear pair.
? ?1 + ?2 = 180º
? 2k + 3k = 180º
? 5k = 180º
? k = 36º
? ?1 = 2k = 2 × 36º = 72º
?2 = 3k = 3 × 36º = 108º
Now,
?3 = ?1 = 72º Verticallyoppositeangles
?4 = ?2 = 108º Verticallyoppositeangles
It is given that, l || m and t is a transversal.
? ?5 = ?1 = 72º Pairofcorrespondingangles
?6 = ?2 = 108º Pairofcorrespondingangles
?7 = ?1 = 72º Pairofalternateexteriorangles
?8 = ?2 = 108º Pairofalternateexteriorangles
Question:32
For what value of x will the line l and m be parallel to each other?
Solution:
For the lines l and m to be parallel
? 3x -20 = 2x +10 [Corresponding Angles] ? x = 30
Question:33
For what value of x will the lines l and m be parallel to each other?
Solution:
? 3x +5 +4x = 180 [Consecutive Interior Angles] ? 7x = 175 ? x = 25
Question:34
In the given figure, AB || CD and BC || ED. Find the value of x.
Solution:
BC ? ED
and CD is the transversal.
Then,
?BCD + ?CDE = 180° [Angles on the same side of a transversal line are supplementary] ? ?BCD +75 = 180 ? ?BCD = 105°
AB ? CD
and BC is the transversal.
?ABC = ?BCD (alternate angles) ? x° = 105° ? x = 105
Question:35
In the given figure, AB || CD || EF. Find the value of x.
Solution:
EF ? CD
and CE is the transversal.
Then,
?ECD + ?CEF = 180° [Consecutive Interior Angles] ? ?ECD +130° = 180° ? ?ECD = 50°
Again, AB ? CD
and BC is the transversal.
Then,
?ABC = ?BCD [Alternate Interior Angles] ? 70° = x +50° [ ? ?BCD = ?BCE + ?ECD] ? x = 20°
Question:36
In the given figure, AB || CD. Find the values of x, y and z.
Solution:
AB ? CD
and let EF and EG be the transversals.
Now, AB ? CD
and EF is the transversal.
Then,
?AEF = ?EFG [Alternate Angles] ? y° = 75° ? y = 75
Also,
?EFC + ?EFD = 180° [Since CFD is a straight line] ? x +y = 180 ? x +75 = 180 ? x = 105
And,
?EGF+ ?EGD = 180° [Since CFGD is a straight line] ? ?EGF+125 = 180 ? ?EGF = 55°
We know that the sum of angles of a triangle is 180°
?EFG+ ?GEF+ ?EGF = 180° ? y +z +55 = 180 ? 75 +z +55 = 180 ? z = 50 ? x = 105, y = 75 and z = 50
Question:37
In each of the figures given below, AB || CD. Find the value of x in each case.
Solution:
i
Draw EF ? AB ? CD
.
Now, AB ? EF
and BE is the transversal.
Then,
?ABE = ?BEF [Alternate Interior Angles] ? ?BEF = 35°
Page 4
Question:29
In the given figure, l || m and a transversal t cuts them. If ?1 = 120°, find the measure of each of the remaining marked angles.
Solution:
We have, ?1 = 120°
. Then,
?1 = ?5 [Corresponding angles] ? ?5 = 120° ?1 = ?3 [Vertically-opposite angles] ? ?3 = 120°
?5 = ?7 [Vertically-opposite angles] ? ?7 = 120° ?1 + ?2 = 180° [Since AFB is a straight line] ? 120°+ ?2 = 180°
? ?2 = 60° ?2 = ?4 [Vertically-opposite angles] ? ?4 = 60° ?2 = ?6 [Corresponding angles]
? ?6 = 60° ?6 = ?8 [Vertically-opposite angles] ? ?8 = 60° ? ?1 = 120°, ?2 = 60°, ?3 = 120°, ?4 = 60°, ?5 = 120°, ?6 = 60°, ?7 = 120° and ?8 = 60°
Question:30
In the given figure, l || m and a transversal t cuts them. If ?7 = 80°, find the measure of each of the remaining marked angles.
Solution:
In the given figure, ?7 and ?8 form a linear pair.
? ?7 + ?8 = 180º
? 80º + ?8 = 180º
? ?8 = 180º - 80º = 100º
Now,
?6 = ?8 = 100º Verticallyoppositeangles
?5 = ?7 = 80º Verticallyoppositeangles
It is given that, l || m and t is a transversal.
? ?1 = ?5 = 80º Pairofcorrespondingangles
?2 = ?6 = 100º Pairofcorrespondingangles
?3 = ?7 = 80º Pairofcorrespondingangles
?4 = ?8 = 100º Pairofcorrespondingangles
Question:31
In the given figure, l || m and a transversal t cuts them. If ?1 : ?2 = 2 : 3, find the measure of each of the marked angles.
Solution:
Let ?1 = 2k and ?2 = 3k, where k is some constant.
Now, ?1 and ?2 form a linear pair.
? ?1 + ?2 = 180º
? 2k + 3k = 180º
? 5k = 180º
? k = 36º
? ?1 = 2k = 2 × 36º = 72º
?2 = 3k = 3 × 36º = 108º
Now,
?3 = ?1 = 72º Verticallyoppositeangles
?4 = ?2 = 108º Verticallyoppositeangles
It is given that, l || m and t is a transversal.
? ?5 = ?1 = 72º Pairofcorrespondingangles
?6 = ?2 = 108º Pairofcorrespondingangles
?7 = ?1 = 72º Pairofalternateexteriorangles
?8 = ?2 = 108º Pairofalternateexteriorangles
Question:32
For what value of x will the line l and m be parallel to each other?
Solution:
For the lines l and m to be parallel
? 3x -20 = 2x +10 [Corresponding Angles] ? x = 30
Question:33
For what value of x will the lines l and m be parallel to each other?
Solution:
? 3x +5 +4x = 180 [Consecutive Interior Angles] ? 7x = 175 ? x = 25
Question:34
In the given figure, AB || CD and BC || ED. Find the value of x.
Solution:
BC ? ED
and CD is the transversal.
Then,
?BCD + ?CDE = 180° [Angles on the same side of a transversal line are supplementary] ? ?BCD +75 = 180 ? ?BCD = 105°
AB ? CD
and BC is the transversal.
?ABC = ?BCD (alternate angles) ? x° = 105° ? x = 105
Question:35
In the given figure, AB || CD || EF. Find the value of x.
Solution:
EF ? CD
and CE is the transversal.
Then,
?ECD + ?CEF = 180° [Consecutive Interior Angles] ? ?ECD +130° = 180° ? ?ECD = 50°
Again, AB ? CD
and BC is the transversal.
Then,
?ABC = ?BCD [Alternate Interior Angles] ? 70° = x +50° [ ? ?BCD = ?BCE + ?ECD] ? x = 20°
Question:36
In the given figure, AB || CD. Find the values of x, y and z.
Solution:
AB ? CD
and let EF and EG be the transversals.
Now, AB ? CD
and EF is the transversal.
Then,
?AEF = ?EFG [Alternate Angles] ? y° = 75° ? y = 75
Also,
?EFC + ?EFD = 180° [Since CFD is a straight line] ? x +y = 180 ? x +75 = 180 ? x = 105
And,
?EGF+ ?EGD = 180° [Since CFGD is a straight line] ? ?EGF+125 = 180 ? ?EGF = 55°
We know that the sum of angles of a triangle is 180°
?EFG+ ?GEF+ ?EGF = 180° ? y +z +55 = 180 ? 75 +z +55 = 180 ? z = 50 ? x = 105, y = 75 and z = 50
Question:37
In each of the figures given below, AB || CD. Find the value of x in each case.
Solution:
i
Draw EF ? AB ? CD
.
Now, AB ? EF
and BE is the transversal.
Then,
?ABE = ?BEF [Alternate Interior Angles] ? ?BEF = 35°
Again, EF ? CD
and DE is the transversal.
Then,
?DEF = ?FED ? ?FED = 65° ? x° = ?BEF+ ?FED = (35 +65)° = 100°or, x = 100
ii
Draw EO ? AB ? CD
.
Then, ?EOB + ?EOD = x°
Now, EO ? AB
and BO is the transversal.
? ?EOB + ?ABO = 180° [Consecutive Interior Angles] ? ?EOB +55° = 180° ? ?EOB = 125°
Again, EO ? CD
and DO is the transversal.
? ?EOD + ?CDO = 180° [Consecutive Interior Angles] ? ?EOD +25° = 180° ? ?EOD = 155°
Therefore,
x° = ?EOB + ?EOD = (125 +155)° = 280°or, x = 280
iii
Draw EF ? AB ? CD
.
Then, ?AEF+ ?CEF = x°
Now, EF ? AB
and AE is the transversal.
? ?AEF+ ?BAE = 180° [Consecutive Interior Angles] ? ?AEF+116 = 180 ? ?AEF = 64°
Again, EF ? CD
and CE is the transversal.
?CEF+ ?ECD = 180° [Consecutive Interior Angles] ? ?CEF+124 = 180 ? ?CEF = 56°
Therefore,
x° = ?AEF+ ?CEF = (64 +56)° = 120°or, x = 120
Question:38
In the given figures, AB || CD. Find the value of x.
Solution:
Draw EF ? AB ? CD
.
EF ? CD
and CE is the transversal.
Then,
?ECD + ?CEF = 180° [Angles on the same side of a transversal line are supplementary] ? 130°+ ?CEF = 180° ? ?CEF = 50°
Again, EF ? AB
and AE is the transversal.
Then,
?BAE + ?AEF = 180° [Angles on the same side of a transversal line are supplementary] ? x°+20°+50° = 180° [ ?AEF = ?AEC + ?CEF] ? x°+70° = 180° ? x° = 110° ? x = 110
Question:39
In the given figure, AB || PQ. Find the values of x and y.
Page 5
Question:29
In the given figure, l || m and a transversal t cuts them. If ?1 = 120°, find the measure of each of the remaining marked angles.
Solution:
We have, ?1 = 120°
. Then,
?1 = ?5 [Corresponding angles] ? ?5 = 120° ?1 = ?3 [Vertically-opposite angles] ? ?3 = 120°
?5 = ?7 [Vertically-opposite angles] ? ?7 = 120° ?1 + ?2 = 180° [Since AFB is a straight line] ? 120°+ ?2 = 180°
? ?2 = 60° ?2 = ?4 [Vertically-opposite angles] ? ?4 = 60° ?2 = ?6 [Corresponding angles]
? ?6 = 60° ?6 = ?8 [Vertically-opposite angles] ? ?8 = 60° ? ?1 = 120°, ?2 = 60°, ?3 = 120°, ?4 = 60°, ?5 = 120°, ?6 = 60°, ?7 = 120° and ?8 = 60°
Question:30
In the given figure, l || m and a transversal t cuts them. If ?7 = 80°, find the measure of each of the remaining marked angles.
Solution:
In the given figure, ?7 and ?8 form a linear pair.
? ?7 + ?8 = 180º
? 80º + ?8 = 180º
? ?8 = 180º - 80º = 100º
Now,
?6 = ?8 = 100º Verticallyoppositeangles
?5 = ?7 = 80º Verticallyoppositeangles
It is given that, l || m and t is a transversal.
? ?1 = ?5 = 80º Pairofcorrespondingangles
?2 = ?6 = 100º Pairofcorrespondingangles
?3 = ?7 = 80º Pairofcorrespondingangles
?4 = ?8 = 100º Pairofcorrespondingangles
Question:31
In the given figure, l || m and a transversal t cuts them. If ?1 : ?2 = 2 : 3, find the measure of each of the marked angles.
Solution:
Let ?1 = 2k and ?2 = 3k, where k is some constant.
Now, ?1 and ?2 form a linear pair.
? ?1 + ?2 = 180º
? 2k + 3k = 180º
? 5k = 180º
? k = 36º
? ?1 = 2k = 2 × 36º = 72º
?2 = 3k = 3 × 36º = 108º
Now,
?3 = ?1 = 72º Verticallyoppositeangles
?4 = ?2 = 108º Verticallyoppositeangles
It is given that, l || m and t is a transversal.
? ?5 = ?1 = 72º Pairofcorrespondingangles
?6 = ?2 = 108º Pairofcorrespondingangles
?7 = ?1 = 72º Pairofalternateexteriorangles
?8 = ?2 = 108º Pairofalternateexteriorangles
Question:32
For what value of x will the line l and m be parallel to each other?
Solution:
For the lines l and m to be parallel
? 3x -20 = 2x +10 [Corresponding Angles] ? x = 30
Question:33
For what value of x will the lines l and m be parallel to each other?
Solution:
? 3x +5 +4x = 180 [Consecutive Interior Angles] ? 7x = 175 ? x = 25
Question:34
In the given figure, AB || CD and BC || ED. Find the value of x.
Solution:
BC ? ED
and CD is the transversal.
Then,
?BCD + ?CDE = 180° [Angles on the same side of a transversal line are supplementary] ? ?BCD +75 = 180 ? ?BCD = 105°
AB ? CD
and BC is the transversal.
?ABC = ?BCD (alternate angles) ? x° = 105° ? x = 105
Question:35
In the given figure, AB || CD || EF. Find the value of x.
Solution:
EF ? CD
and CE is the transversal.
Then,
?ECD + ?CEF = 180° [Consecutive Interior Angles] ? ?ECD +130° = 180° ? ?ECD = 50°
Again, AB ? CD
and BC is the transversal.
Then,
?ABC = ?BCD [Alternate Interior Angles] ? 70° = x +50° [ ? ?BCD = ?BCE + ?ECD] ? x = 20°
Question:36
In the given figure, AB || CD. Find the values of x, y and z.
Solution:
AB ? CD
and let EF and EG be the transversals.
Now, AB ? CD
and EF is the transversal.
Then,
?AEF = ?EFG [Alternate Angles] ? y° = 75° ? y = 75
Also,
?EFC + ?EFD = 180° [Since CFD is a straight line] ? x +y = 180 ? x +75 = 180 ? x = 105
And,
?EGF+ ?EGD = 180° [Since CFGD is a straight line] ? ?EGF+125 = 180 ? ?EGF = 55°
We know that the sum of angles of a triangle is 180°
?EFG+ ?GEF+ ?EGF = 180° ? y +z +55 = 180 ? 75 +z +55 = 180 ? z = 50 ? x = 105, y = 75 and z = 50
Question:37
In each of the figures given below, AB || CD. Find the value of x in each case.
Solution:
i
Draw EF ? AB ? CD
.
Now, AB ? EF
and BE is the transversal.
Then,
?ABE = ?BEF [Alternate Interior Angles] ? ?BEF = 35°
Again, EF ? CD
and DE is the transversal.
Then,
?DEF = ?FED ? ?FED = 65° ? x° = ?BEF+ ?FED = (35 +65)° = 100°or, x = 100
ii
Draw EO ? AB ? CD
.
Then, ?EOB + ?EOD = x°
Now, EO ? AB
and BO is the transversal.
? ?EOB + ?ABO = 180° [Consecutive Interior Angles] ? ?EOB +55° = 180° ? ?EOB = 125°
Again, EO ? CD
and DO is the transversal.
? ?EOD + ?CDO = 180° [Consecutive Interior Angles] ? ?EOD +25° = 180° ? ?EOD = 155°
Therefore,
x° = ?EOB + ?EOD = (125 +155)° = 280°or, x = 280
iii
Draw EF ? AB ? CD
.
Then, ?AEF+ ?CEF = x°
Now, EF ? AB
and AE is the transversal.
? ?AEF+ ?BAE = 180° [Consecutive Interior Angles] ? ?AEF+116 = 180 ? ?AEF = 64°
Again, EF ? CD
and CE is the transversal.
?CEF+ ?ECD = 180° [Consecutive Interior Angles] ? ?CEF+124 = 180 ? ?CEF = 56°
Therefore,
x° = ?AEF+ ?CEF = (64 +56)° = 120°or, x = 120
Question:38
In the given figures, AB || CD. Find the value of x.
Solution:
Draw EF ? AB ? CD
.
EF ? CD
and CE is the transversal.
Then,
?ECD + ?CEF = 180° [Angles on the same side of a transversal line are supplementary] ? 130°+ ?CEF = 180° ? ?CEF = 50°
Again, EF ? AB
and AE is the transversal.
Then,
?BAE + ?AEF = 180° [Angles on the same side of a transversal line are supplementary] ? x°+20°+50° = 180° [ ?AEF = ?AEC + ?CEF] ? x°+70° = 180° ? x° = 110° ? x = 110
Question:39
In the given figure, AB || PQ. Find the values of x and y.
Solution:
Given, AB ? PQ
.
Let CD be the transversal cutting AB and PQ at E and F, respectively.
Then,
?CEB + ?BEG+ ?GEF = 180° [Since CD is a straight line] ? 75°+20°+ ?GEF = 180° ? ?GEF = 85°
We know that the sum of angles of a triangle is 180°
.
? ?GEF+ ?EGF+ ?EFG = 180 ? 85°+x +25° = 180° ? 110°+x = 180° ? x = 70°
And
?FEG+ ?BEG = ?DFQ [Corresponding Angles] ? 85°+20° = ?DFQ ? ?DFQ = 105° ?EFG+ ?GFQ+ ?DFQ = 180° [Since CD is a straight line] ? 25°+y +105° = 180° ? y = 50° ? x =
Question:40
In the given figure, AB || CD. Find the value of x.
Solution:
AB ? CD
and AC is the transversal.
Then,
?BAC + ?ACD = 180° [Consecutive Interior Angles] ? 75 + ?ACD = 180 ? ?ACD = 105°
And,
?ACD = ?ECF [Vertically-Opposite Angles] ? ?ECF = 105°
We know that the sum of the angles of a triangle is 180°
.
?ECF+ ?CFE + ?CEF = 180° ? 105°+30°+x = 180° ? 135°+x = 180° ? x = 45°
Question:41
In the given figure, AB || CD. Find the value of x.
Solution:
AB ? CD
and PQ is the transversal.
Then,
?PEF = ?EGH [Corresponding Angles] ? ?EGH = 85°
And,
?EGH + ?QGH = 180° [Since PQ is a straight line] ? 85°+ ?QGH = 180° ? ?QGH = 95°
Also,
?CHQ+ ?GHQ = 180° [Since CD is a straight line] ? 115°+ ?GHQ = 180° ? ?GHQ = 65°
We know that the sum of angles of a triangle is 180°
.
? ?QGH + ?GHQ+ ?GQH = 180° ? 95°+65°+x = 180° ? x = 20° ? x = 20°
Question:42
In the given figure, AB || CD. Find the values of x, y and z.
Solution:
?ADC = ?DAB [Alternate Interior Angles] ? z = 75° ?ABC = ?BCD [Alternate Interior Angles] ? x = 35°
We know that the sum of the angles of a triangle is 180°
.
? 35°+y +75° = 180° ? y = 70° ? x = 35°, y = 70° and z = 75°.
Question:43
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