Page 1
Question:13
In the adjoining figure, AOB is a straight line. Find the value of x.
Solution:
We know that the sum of angles in a linear pair is 180°
.
Therefore,
?AOC + ?BOC = 180° ? 62°+x° = 180° ? x° = (180°-62°) ? x = 118°
Hence, the value of x is 118°
.
Question:14
In the adjoining figure, AOB is a straight line. Find the value of x. Hence, find ?AOC and ?BOD.
Solution:
As AOB is a straight line, the sum of angles on the same side of AOB, at a point O on it, is 180°
.
Therefore,
?AOC + ?COD + ?BOD = 180° ? (3x -7)°+55°+(x +20)° = 180 ? 4x = 112° ? x = 28°
Hence,
?AOC = 3x -7
= 3 ×28 -7 = 77°
and ?BOD = x +20
= 28 +20 = 48°
Question:15
In the adjoining figure, AOB is a straight line. Find the value of x. Also, find ?AOC, ?COD and ?BOD.
Solution:
AOB is a straight line. Therefore,
?AOC + ?COD + ?BOD = 180° ? (3x +7)°+(2x -19)°+x° = 180° ? 6x = 192° ? x = 32°
Therefore,
?AOC = 3 ×32°+7 = 103° ?COD = 2 ×32°-19 = 45° and ?BOD = 32°
Question:16
In the adjoining figure, x:y:z = 5:4:6. If XOY is a straight line, find the values of x, y and z.
Solution:
Let x = 5a, y = 4a and z = 6a
XOY is a straight line. Therefore,
?XOP + ?POQ+ ?YOQ = 180° ? 5a +4a +6a = 180° ? 15a = 180° ? a = 12°
Therefore,
x ? 5 ×12° = 60° y ? 4 ×12° = 48°and z ? 6 ×12° = 72°
Question:17
In the adjoining figure, what value of x will make AOB a straight line?
Page 2
Question:13
In the adjoining figure, AOB is a straight line. Find the value of x.
Solution:
We know that the sum of angles in a linear pair is 180°
.
Therefore,
?AOC + ?BOC = 180° ? 62°+x° = 180° ? x° = (180°-62°) ? x = 118°
Hence, the value of x is 118°
.
Question:14
In the adjoining figure, AOB is a straight line. Find the value of x. Hence, find ?AOC and ?BOD.
Solution:
As AOB is a straight line, the sum of angles on the same side of AOB, at a point O on it, is 180°
.
Therefore,
?AOC + ?COD + ?BOD = 180° ? (3x -7)°+55°+(x +20)° = 180 ? 4x = 112° ? x = 28°
Hence,
?AOC = 3x -7
= 3 ×28 -7 = 77°
and ?BOD = x +20
= 28 +20 = 48°
Question:15
In the adjoining figure, AOB is a straight line. Find the value of x. Also, find ?AOC, ?COD and ?BOD.
Solution:
AOB is a straight line. Therefore,
?AOC + ?COD + ?BOD = 180° ? (3x +7)°+(2x -19)°+x° = 180° ? 6x = 192° ? x = 32°
Therefore,
?AOC = 3 ×32°+7 = 103° ?COD = 2 ×32°-19 = 45° and ?BOD = 32°
Question:16
In the adjoining figure, x:y:z = 5:4:6. If XOY is a straight line, find the values of x, y and z.
Solution:
Let x = 5a, y = 4a and z = 6a
XOY is a straight line. Therefore,
?XOP + ?POQ+ ?YOQ = 180° ? 5a +4a +6a = 180° ? 15a = 180° ? a = 12°
Therefore,
x ? 5 ×12° = 60° y ? 4 ×12° = 48°and z ? 6 ×12° = 72°
Question:17
In the adjoining figure, what value of x will make AOB a straight line?
Solution:
AOB will be a straight line if
3x +20 +4x -36 = 180° ? 7x = 196° ? x = 28°
Hence, x = 28 will make AOB a straight line.
Question:18
Two lines AB and CD intersect at O. If ?AOC = 50°, find ?AOD, ?BOD and ?BOC.
Solution:
We know that if two lines intersect then the vertically-opposite angles are equal.
Therefore, ?AOC = ?BOD = 50°
Let ?AOD = ?BOC = x°
Also, we know that the sum of all angles around a point is 360°
.
Therefore,
?AOC + ?AOD + ?BOD + ?BOC = 360° ? 50 +x +50 +x = 360° ? 2x = 260° ? x = 130°
Hence, ?AOD = ?BOC = 130°
Therefore, ?AOD = 130°, ?BOD = 50° and ?BOC = 130°
.
Question:19
In the adjoining figure, three coplanar lines AB, CD and EF intersect at a point O, forming angles as shown. Find the values of x, y, z and t.
Solution:
We know that if two lines intersect, then the vertically opposite angles are equal.
? ?BOD = ?AOC = 90°
Hence, t = 90°
Also,
?DOF = ?COE = 50°
Hence, z = 50°
Since, AOB is a straight line, we have:
?AOC + ?COE + ?BOE = 180° ? 90 +50 +y = 180° ? 140 +y = 180° ? y = 40°
Also,
?BOE = ?AOF = 40°
Hence, x = 40°
? x = 40°, y = 40°, z = 50° and t = 90°
Question:20
In the adjoining figure, three coplanar lines AB, CD and EF intersect at a point O. Find the value of x. Also, find ?AOD, ?COE and ?AOE.
Solution:
We know that if two lines intersect, then the vertically-opposite angles are equal.
? ?DOF = ?COE = 5x° ?AOD = ?BOC = 2x° and ?AOE = ?BOF = 3x°
Since, AOB is a straight line, we have:
?AOE + ?COE + ?BOC = 180° ? 3x +5x +2x = 180° ? 10x = 180° ? x = 18°
Therefore,
?AOD = 2 ×18° = 36° ?COE = 5 ×18° = 90° ?AOE = 3 ×18° = 54°
Question:21
Two adjacent angles on a straight line are in the ratio 5 : 4. Find the measure of each of these angles.
Solution:
Let the two adjacent angles be 5x and 4x, respectively.
Then,
5x +4x = 180° ? 9x = 180° ? x = 20°
Page 3
Question:13
In the adjoining figure, AOB is a straight line. Find the value of x.
Solution:
We know that the sum of angles in a linear pair is 180°
.
Therefore,
?AOC + ?BOC = 180° ? 62°+x° = 180° ? x° = (180°-62°) ? x = 118°
Hence, the value of x is 118°
.
Question:14
In the adjoining figure, AOB is a straight line. Find the value of x. Hence, find ?AOC and ?BOD.
Solution:
As AOB is a straight line, the sum of angles on the same side of AOB, at a point O on it, is 180°
.
Therefore,
?AOC + ?COD + ?BOD = 180° ? (3x -7)°+55°+(x +20)° = 180 ? 4x = 112° ? x = 28°
Hence,
?AOC = 3x -7
= 3 ×28 -7 = 77°
and ?BOD = x +20
= 28 +20 = 48°
Question:15
In the adjoining figure, AOB is a straight line. Find the value of x. Also, find ?AOC, ?COD and ?BOD.
Solution:
AOB is a straight line. Therefore,
?AOC + ?COD + ?BOD = 180° ? (3x +7)°+(2x -19)°+x° = 180° ? 6x = 192° ? x = 32°
Therefore,
?AOC = 3 ×32°+7 = 103° ?COD = 2 ×32°-19 = 45° and ?BOD = 32°
Question:16
In the adjoining figure, x:y:z = 5:4:6. If XOY is a straight line, find the values of x, y and z.
Solution:
Let x = 5a, y = 4a and z = 6a
XOY is a straight line. Therefore,
?XOP + ?POQ+ ?YOQ = 180° ? 5a +4a +6a = 180° ? 15a = 180° ? a = 12°
Therefore,
x ? 5 ×12° = 60° y ? 4 ×12° = 48°and z ? 6 ×12° = 72°
Question:17
In the adjoining figure, what value of x will make AOB a straight line?
Solution:
AOB will be a straight line if
3x +20 +4x -36 = 180° ? 7x = 196° ? x = 28°
Hence, x = 28 will make AOB a straight line.
Question:18
Two lines AB and CD intersect at O. If ?AOC = 50°, find ?AOD, ?BOD and ?BOC.
Solution:
We know that if two lines intersect then the vertically-opposite angles are equal.
Therefore, ?AOC = ?BOD = 50°
Let ?AOD = ?BOC = x°
Also, we know that the sum of all angles around a point is 360°
.
Therefore,
?AOC + ?AOD + ?BOD + ?BOC = 360° ? 50 +x +50 +x = 360° ? 2x = 260° ? x = 130°
Hence, ?AOD = ?BOC = 130°
Therefore, ?AOD = 130°, ?BOD = 50° and ?BOC = 130°
.
Question:19
In the adjoining figure, three coplanar lines AB, CD and EF intersect at a point O, forming angles as shown. Find the values of x, y, z and t.
Solution:
We know that if two lines intersect, then the vertically opposite angles are equal.
? ?BOD = ?AOC = 90°
Hence, t = 90°
Also,
?DOF = ?COE = 50°
Hence, z = 50°
Since, AOB is a straight line, we have:
?AOC + ?COE + ?BOE = 180° ? 90 +50 +y = 180° ? 140 +y = 180° ? y = 40°
Also,
?BOE = ?AOF = 40°
Hence, x = 40°
? x = 40°, y = 40°, z = 50° and t = 90°
Question:20
In the adjoining figure, three coplanar lines AB, CD and EF intersect at a point O. Find the value of x. Also, find ?AOD, ?COE and ?AOE.
Solution:
We know that if two lines intersect, then the vertically-opposite angles are equal.
? ?DOF = ?COE = 5x° ?AOD = ?BOC = 2x° and ?AOE = ?BOF = 3x°
Since, AOB is a straight line, we have:
?AOE + ?COE + ?BOC = 180° ? 3x +5x +2x = 180° ? 10x = 180° ? x = 18°
Therefore,
?AOD = 2 ×18° = 36° ?COE = 5 ×18° = 90° ?AOE = 3 ×18° = 54°
Question:21
Two adjacent angles on a straight line are in the ratio 5 : 4. Find the measure of each of these angles.
Solution:
Let the two adjacent angles be 5x and 4x, respectively.
Then,
5x +4x = 180° ? 9x = 180° ? x = 20°
Hence, the two angles are 5 ×20° = 100° and 4 ×20° = 80°
.
Question:22
If two straight lines intersect in such a way that one of the angles formed measures 90°, show that each of the remaining angles measures 90°.
Solution:
We know that if two lines intersect, then the vertically-opposite angles are equal.
?AOC = 90°. Then, ?AOC = ?BOD = 90°.
And let ?BOC = ?AOD = x
Also, we know that the sum of all angles around a point is 360°
? ?AOC + ?BOD + ?AOD + ?BOC = 360° ? 90°+90°+x +x = 360° ? 2x = 180° ? x = 90°
Hence, ?BOC = ?AOD = 90°
? ?AOC = ?BOD = ?BOC = ?AOD = 90°
Hence, the measure of each of the remaining angles is 90
o
.
Question:23
Two lines AB and CD intersect at a point O, such that ?BOC + ?AOD = 280°, as shown in the figure. Find all the four angles.
Solution:
We know that if two lines intersect, then the vertically-opposite angles are equal.
Let ?BOC = ?AOD = x°
Then,
x +x = 280 ? 2x = 280 ? x = 140° ? ?BOC = ?AOD = 140°
Also, let ?AOC = ?BOD = y°
We know that the sum of all angles around a point is 360°
.
? ?AOC + ?BOC + ?BOD + ?AOD = 360° ? y +140 +y +140 = 360° ? 2y = 80° ? y = 40°
Hence, ?AOC = ?BOD = 40°
? ?BOC = ?AOD = 140° and ?AOC = ?BOD = 40°
Question:24
Two lines AB and CD intersect each other at a point O such that ?AOC : ?AOD = 5 : 7. Find all the angles.
Solution:
Let ?AOC = 5k and ?AOD = 7k, where k is some constant.
Here, ?AOC and ?AOD form a linear pair.
? ?AOC + ?AOD = 180º
? 5k + 7k = 180º
? 12k = 180º
? k = 15º
? ?AOC = 5k = 5 × 15º = 75º
?AOD = 7k = 7 × 15º = 105º
Now, ?BOD = ?AOC = 75º Verticallyoppositeangles
?BOC = ?AOD = 105º Verticallyoppositeangles
Question:25
In the given figure, three lines AB, CD and EF intersect at a point O such that ?AOE = 35° and ?BOD = 40°. Find the measure of ?AOC, ?BOF, ?COF and ?DOE.
Page 4
Question:13
In the adjoining figure, AOB is a straight line. Find the value of x.
Solution:
We know that the sum of angles in a linear pair is 180°
.
Therefore,
?AOC + ?BOC = 180° ? 62°+x° = 180° ? x° = (180°-62°) ? x = 118°
Hence, the value of x is 118°
.
Question:14
In the adjoining figure, AOB is a straight line. Find the value of x. Hence, find ?AOC and ?BOD.
Solution:
As AOB is a straight line, the sum of angles on the same side of AOB, at a point O on it, is 180°
.
Therefore,
?AOC + ?COD + ?BOD = 180° ? (3x -7)°+55°+(x +20)° = 180 ? 4x = 112° ? x = 28°
Hence,
?AOC = 3x -7
= 3 ×28 -7 = 77°
and ?BOD = x +20
= 28 +20 = 48°
Question:15
In the adjoining figure, AOB is a straight line. Find the value of x. Also, find ?AOC, ?COD and ?BOD.
Solution:
AOB is a straight line. Therefore,
?AOC + ?COD + ?BOD = 180° ? (3x +7)°+(2x -19)°+x° = 180° ? 6x = 192° ? x = 32°
Therefore,
?AOC = 3 ×32°+7 = 103° ?COD = 2 ×32°-19 = 45° and ?BOD = 32°
Question:16
In the adjoining figure, x:y:z = 5:4:6. If XOY is a straight line, find the values of x, y and z.
Solution:
Let x = 5a, y = 4a and z = 6a
XOY is a straight line. Therefore,
?XOP + ?POQ+ ?YOQ = 180° ? 5a +4a +6a = 180° ? 15a = 180° ? a = 12°
Therefore,
x ? 5 ×12° = 60° y ? 4 ×12° = 48°and z ? 6 ×12° = 72°
Question:17
In the adjoining figure, what value of x will make AOB a straight line?
Solution:
AOB will be a straight line if
3x +20 +4x -36 = 180° ? 7x = 196° ? x = 28°
Hence, x = 28 will make AOB a straight line.
Question:18
Two lines AB and CD intersect at O. If ?AOC = 50°, find ?AOD, ?BOD and ?BOC.
Solution:
We know that if two lines intersect then the vertically-opposite angles are equal.
Therefore, ?AOC = ?BOD = 50°
Let ?AOD = ?BOC = x°
Also, we know that the sum of all angles around a point is 360°
.
Therefore,
?AOC + ?AOD + ?BOD + ?BOC = 360° ? 50 +x +50 +x = 360° ? 2x = 260° ? x = 130°
Hence, ?AOD = ?BOC = 130°
Therefore, ?AOD = 130°, ?BOD = 50° and ?BOC = 130°
.
Question:19
In the adjoining figure, three coplanar lines AB, CD and EF intersect at a point O, forming angles as shown. Find the values of x, y, z and t.
Solution:
We know that if two lines intersect, then the vertically opposite angles are equal.
? ?BOD = ?AOC = 90°
Hence, t = 90°
Also,
?DOF = ?COE = 50°
Hence, z = 50°
Since, AOB is a straight line, we have:
?AOC + ?COE + ?BOE = 180° ? 90 +50 +y = 180° ? 140 +y = 180° ? y = 40°
Also,
?BOE = ?AOF = 40°
Hence, x = 40°
? x = 40°, y = 40°, z = 50° and t = 90°
Question:20
In the adjoining figure, three coplanar lines AB, CD and EF intersect at a point O. Find the value of x. Also, find ?AOD, ?COE and ?AOE.
Solution:
We know that if two lines intersect, then the vertically-opposite angles are equal.
? ?DOF = ?COE = 5x° ?AOD = ?BOC = 2x° and ?AOE = ?BOF = 3x°
Since, AOB is a straight line, we have:
?AOE + ?COE + ?BOC = 180° ? 3x +5x +2x = 180° ? 10x = 180° ? x = 18°
Therefore,
?AOD = 2 ×18° = 36° ?COE = 5 ×18° = 90° ?AOE = 3 ×18° = 54°
Question:21
Two adjacent angles on a straight line are in the ratio 5 : 4. Find the measure of each of these angles.
Solution:
Let the two adjacent angles be 5x and 4x, respectively.
Then,
5x +4x = 180° ? 9x = 180° ? x = 20°
Hence, the two angles are 5 ×20° = 100° and 4 ×20° = 80°
.
Question:22
If two straight lines intersect in such a way that one of the angles formed measures 90°, show that each of the remaining angles measures 90°.
Solution:
We know that if two lines intersect, then the vertically-opposite angles are equal.
?AOC = 90°. Then, ?AOC = ?BOD = 90°.
And let ?BOC = ?AOD = x
Also, we know that the sum of all angles around a point is 360°
? ?AOC + ?BOD + ?AOD + ?BOC = 360° ? 90°+90°+x +x = 360° ? 2x = 180° ? x = 90°
Hence, ?BOC = ?AOD = 90°
? ?AOC = ?BOD = ?BOC = ?AOD = 90°
Hence, the measure of each of the remaining angles is 90
o
.
Question:23
Two lines AB and CD intersect at a point O, such that ?BOC + ?AOD = 280°, as shown in the figure. Find all the four angles.
Solution:
We know that if two lines intersect, then the vertically-opposite angles are equal.
Let ?BOC = ?AOD = x°
Then,
x +x = 280 ? 2x = 280 ? x = 140° ? ?BOC = ?AOD = 140°
Also, let ?AOC = ?BOD = y°
We know that the sum of all angles around a point is 360°
.
? ?AOC + ?BOC + ?BOD + ?AOD = 360° ? y +140 +y +140 = 360° ? 2y = 80° ? y = 40°
Hence, ?AOC = ?BOD = 40°
? ?BOC = ?AOD = 140° and ?AOC = ?BOD = 40°
Question:24
Two lines AB and CD intersect each other at a point O such that ?AOC : ?AOD = 5 : 7. Find all the angles.
Solution:
Let ?AOC = 5k and ?AOD = 7k, where k is some constant.
Here, ?AOC and ?AOD form a linear pair.
? ?AOC + ?AOD = 180º
? 5k + 7k = 180º
? 12k = 180º
? k = 15º
? ?AOC = 5k = 5 × 15º = 75º
?AOD = 7k = 7 × 15º = 105º
Now, ?BOD = ?AOC = 75º Verticallyoppositeangles
?BOC = ?AOD = 105º Verticallyoppositeangles
Question:25
In the given figure, three lines AB, CD and EF intersect at a point O such that ?AOE = 35° and ?BOD = 40°. Find the measure of ?AOC, ?BOF, ?COF and ?DOE.
Solution:
In the given figure,
?AOC = ?BOD = 40º Verticallyoppositeangles
?BOF = ?AOE = 35º Verticallyoppositeangles
Now, ?EOC and ?COF form a linear pair.
? ?EOC + ?COF = 180º
? ?AOE + ?AOC
+ ?COF = 180º
? 35º + 40º + ?COF = 180º
? 75º + ?COF = 180º
? ?COF = 180º - 75º = 105º
Also, ?DOE = ?COF = 105º Verticallyoppositeangles
Question:26
In the given figure, the two lines AB and CD intersect at a point O such that ?BOC = 125°. Find the values of x, y and z.
Solution:
Here, ?AOC and ?BOC form a linear pair.
? ?AOC + ?BOC = 180º
? xº + 125º = 180º
? xº = 180º - 125º ? = 55º ? Now,
?AOD = ?BOC = 125º Verticallyoppositeangles
? yº = 125º ? ?BOD = ?AOC = 55º Verticallyoppositeangles
? zº = 55º ? Thus, the respective values of x, y and z are 55, 125 and 55.
Question:27
If two straight lines intersect each other, then prove that the ray opposite the bisector of one of the angles so formed bisects the vertically-opposite angle.
Solution:
Let AB and CD be the two lines intersecting at a point O and let ray OE bisect ?AOC
. Now, draw a ray OF in the opposite direction of OE, such that EOF is a straight line.
Let ?COE = 1, ?AOE = 2, ?BOF = 3 and ?DOF = 4
.
We know that vertically-opposite angles are equal.
? ?1 = ?4 and ?2 = ?3
But, ?1 = ?2
[Since OE bisects ?AOC
]
? ?4 = ?3
Hence, the ray opposite the bisector of one of the angles so formed bisects the vertically-opposite angle.
Question:28
Prove that the bisectors of two adjacent supplementary angles include a right angle.
Solution:
Page 5
Question:13
In the adjoining figure, AOB is a straight line. Find the value of x.
Solution:
We know that the sum of angles in a linear pair is 180°
.
Therefore,
?AOC + ?BOC = 180° ? 62°+x° = 180° ? x° = (180°-62°) ? x = 118°
Hence, the value of x is 118°
.
Question:14
In the adjoining figure, AOB is a straight line. Find the value of x. Hence, find ?AOC and ?BOD.
Solution:
As AOB is a straight line, the sum of angles on the same side of AOB, at a point O on it, is 180°
.
Therefore,
?AOC + ?COD + ?BOD = 180° ? (3x -7)°+55°+(x +20)° = 180 ? 4x = 112° ? x = 28°
Hence,
?AOC = 3x -7
= 3 ×28 -7 = 77°
and ?BOD = x +20
= 28 +20 = 48°
Question:15
In the adjoining figure, AOB is a straight line. Find the value of x. Also, find ?AOC, ?COD and ?BOD.
Solution:
AOB is a straight line. Therefore,
?AOC + ?COD + ?BOD = 180° ? (3x +7)°+(2x -19)°+x° = 180° ? 6x = 192° ? x = 32°
Therefore,
?AOC = 3 ×32°+7 = 103° ?COD = 2 ×32°-19 = 45° and ?BOD = 32°
Question:16
In the adjoining figure, x:y:z = 5:4:6. If XOY is a straight line, find the values of x, y and z.
Solution:
Let x = 5a, y = 4a and z = 6a
XOY is a straight line. Therefore,
?XOP + ?POQ+ ?YOQ = 180° ? 5a +4a +6a = 180° ? 15a = 180° ? a = 12°
Therefore,
x ? 5 ×12° = 60° y ? 4 ×12° = 48°and z ? 6 ×12° = 72°
Question:17
In the adjoining figure, what value of x will make AOB a straight line?
Solution:
AOB will be a straight line if
3x +20 +4x -36 = 180° ? 7x = 196° ? x = 28°
Hence, x = 28 will make AOB a straight line.
Question:18
Two lines AB and CD intersect at O. If ?AOC = 50°, find ?AOD, ?BOD and ?BOC.
Solution:
We know that if two lines intersect then the vertically-opposite angles are equal.
Therefore, ?AOC = ?BOD = 50°
Let ?AOD = ?BOC = x°
Also, we know that the sum of all angles around a point is 360°
.
Therefore,
?AOC + ?AOD + ?BOD + ?BOC = 360° ? 50 +x +50 +x = 360° ? 2x = 260° ? x = 130°
Hence, ?AOD = ?BOC = 130°
Therefore, ?AOD = 130°, ?BOD = 50° and ?BOC = 130°
.
Question:19
In the adjoining figure, three coplanar lines AB, CD and EF intersect at a point O, forming angles as shown. Find the values of x, y, z and t.
Solution:
We know that if two lines intersect, then the vertically opposite angles are equal.
? ?BOD = ?AOC = 90°
Hence, t = 90°
Also,
?DOF = ?COE = 50°
Hence, z = 50°
Since, AOB is a straight line, we have:
?AOC + ?COE + ?BOE = 180° ? 90 +50 +y = 180° ? 140 +y = 180° ? y = 40°
Also,
?BOE = ?AOF = 40°
Hence, x = 40°
? x = 40°, y = 40°, z = 50° and t = 90°
Question:20
In the adjoining figure, three coplanar lines AB, CD and EF intersect at a point O. Find the value of x. Also, find ?AOD, ?COE and ?AOE.
Solution:
We know that if two lines intersect, then the vertically-opposite angles are equal.
? ?DOF = ?COE = 5x° ?AOD = ?BOC = 2x° and ?AOE = ?BOF = 3x°
Since, AOB is a straight line, we have:
?AOE + ?COE + ?BOC = 180° ? 3x +5x +2x = 180° ? 10x = 180° ? x = 18°
Therefore,
?AOD = 2 ×18° = 36° ?COE = 5 ×18° = 90° ?AOE = 3 ×18° = 54°
Question:21
Two adjacent angles on a straight line are in the ratio 5 : 4. Find the measure of each of these angles.
Solution:
Let the two adjacent angles be 5x and 4x, respectively.
Then,
5x +4x = 180° ? 9x = 180° ? x = 20°
Hence, the two angles are 5 ×20° = 100° and 4 ×20° = 80°
.
Question:22
If two straight lines intersect in such a way that one of the angles formed measures 90°, show that each of the remaining angles measures 90°.
Solution:
We know that if two lines intersect, then the vertically-opposite angles are equal.
?AOC = 90°. Then, ?AOC = ?BOD = 90°.
And let ?BOC = ?AOD = x
Also, we know that the sum of all angles around a point is 360°
? ?AOC + ?BOD + ?AOD + ?BOC = 360° ? 90°+90°+x +x = 360° ? 2x = 180° ? x = 90°
Hence, ?BOC = ?AOD = 90°
? ?AOC = ?BOD = ?BOC = ?AOD = 90°
Hence, the measure of each of the remaining angles is 90
o
.
Question:23
Two lines AB and CD intersect at a point O, such that ?BOC + ?AOD = 280°, as shown in the figure. Find all the four angles.
Solution:
We know that if two lines intersect, then the vertically-opposite angles are equal.
Let ?BOC = ?AOD = x°
Then,
x +x = 280 ? 2x = 280 ? x = 140° ? ?BOC = ?AOD = 140°
Also, let ?AOC = ?BOD = y°
We know that the sum of all angles around a point is 360°
.
? ?AOC + ?BOC + ?BOD + ?AOD = 360° ? y +140 +y +140 = 360° ? 2y = 80° ? y = 40°
Hence, ?AOC = ?BOD = 40°
? ?BOC = ?AOD = 140° and ?AOC = ?BOD = 40°
Question:24
Two lines AB and CD intersect each other at a point O such that ?AOC : ?AOD = 5 : 7. Find all the angles.
Solution:
Let ?AOC = 5k and ?AOD = 7k, where k is some constant.
Here, ?AOC and ?AOD form a linear pair.
? ?AOC + ?AOD = 180º
? 5k + 7k = 180º
? 12k = 180º
? k = 15º
? ?AOC = 5k = 5 × 15º = 75º
?AOD = 7k = 7 × 15º = 105º
Now, ?BOD = ?AOC = 75º Verticallyoppositeangles
?BOC = ?AOD = 105º Verticallyoppositeangles
Question:25
In the given figure, three lines AB, CD and EF intersect at a point O such that ?AOE = 35° and ?BOD = 40°. Find the measure of ?AOC, ?BOF, ?COF and ?DOE.
Solution:
In the given figure,
?AOC = ?BOD = 40º Verticallyoppositeangles
?BOF = ?AOE = 35º Verticallyoppositeangles
Now, ?EOC and ?COF form a linear pair.
? ?EOC + ?COF = 180º
? ?AOE + ?AOC
+ ?COF = 180º
? 35º + 40º + ?COF = 180º
? 75º + ?COF = 180º
? ?COF = 180º - 75º = 105º
Also, ?DOE = ?COF = 105º Verticallyoppositeangles
Question:26
In the given figure, the two lines AB and CD intersect at a point O such that ?BOC = 125°. Find the values of x, y and z.
Solution:
Here, ?AOC and ?BOC form a linear pair.
? ?AOC + ?BOC = 180º
? xº + 125º = 180º
? xº = 180º - 125º ? = 55º ? Now,
?AOD = ?BOC = 125º Verticallyoppositeangles
? yº = 125º ? ?BOD = ?AOC = 55º Verticallyoppositeangles
? zº = 55º ? Thus, the respective values of x, y and z are 55, 125 and 55.
Question:27
If two straight lines intersect each other, then prove that the ray opposite the bisector of one of the angles so formed bisects the vertically-opposite angle.
Solution:
Let AB and CD be the two lines intersecting at a point O and let ray OE bisect ?AOC
. Now, draw a ray OF in the opposite direction of OE, such that EOF is a straight line.
Let ?COE = 1, ?AOE = 2, ?BOF = 3 and ?DOF = 4
.
We know that vertically-opposite angles are equal.
? ?1 = ?4 and ?2 = ?3
But, ?1 = ?2
[Since OE bisects ?AOC
]
? ?4 = ?3
Hence, the ray opposite the bisector of one of the angles so formed bisects the vertically-opposite angle.
Question:28
Prove that the bisectors of two adjacent supplementary angles include a right angle.
Solution:
Let AOB denote a straight line and let ?AOC and ?BOC
be the supplementary angles.
Then, we have:
?AOC = x° and ?BOC = (180 -x)°
Let OE bisect ?AOC and OF bisect ?BOC
.
Then, we have:
?AOE = ?COE =
1
2
x° and ?BOF = ?FOC =
1
2
(180 -x)°
Therefore,
?COE + ?FOC =
1
2
x +
1
2
(180°-x)
=
1
2
(x +180°-x) =
1
2
(180°) = 90°
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