Class 9 Exam > Class 9 Notes > Mathematics (Maths) Class 9 > RS Aggarwal Solutions: Lines and Angles- 2
RS Aggarwal Solutions: Lines and Angles- 2 | Mathematics (Maths) Class 9 PDF Download
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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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FAQs on RS Aggarwal Solutions: Lines and Angles- 2 - Mathematics (Maths) Class 9
| 1. What are the different types of angles? |  |
Ans. There are several types of angles, including acute angles (less than 90 degrees), right angles (exactly 90 degrees), obtuse angles (between 90 and 180 degrees), and straight angles (exactly 180 degrees).
| 2. How do you identify parallel lines? | |
Ans. Parallel lines are lines that never intersect and are always the same distance apart. To identify parallel lines, you can use a ruler to measure the distance between the lines at different points. If the distance remains the same, the lines are parallel.
| 3. What is the difference between perpendicular and intersecting lines? | |
Ans. Perpendicular lines are lines that intersect at a right angle (90 degrees), forming a square corner. Intersecting lines, on the other hand, are lines that cross each other at any angle, forming an "X" shape.
| 4. How do you calculate the sum of the interior angles of a polygon? | |
Ans. The sum of the interior angles of a polygon can be calculated using the formula: (n-2) x 180 degrees, where "n" represents the number of sides or vertices of the polygon.
| 5. Can two lines intersect at more than one point? | |
Ans. No, two lines can only intersect at one point. If two lines intersect at more than one point, they are not considered lines but rather overlapping or coinciding lines.
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