## two right angles are always supplementary to each other

⇒ ∠QUR = 180° – 42° = 138°. Let each angle be x. Solution: In which of the following figures, a and bare forming a pair of adjacent angles? Thus, one angle is 89° and other is 91°, Question 99. How do you think about the answers? Now, p || q and m is a transversal. Solution: (c) ∠3 + ∠8 = 180° ∴ 4c = 120°      [Corresponding angles] Those Adjacent Angles Are Complementary. (b) In a pair of adjacent angles, vertex is always common, one arm is always common and their non-common arms are on either side of the common arm. Since, l || m and AB is a transversal. Question 90. (a) supplementary Determine the values of x and y. ⇒ b = 180° – 132° = 48° ∴ Its supplement = 180° – x Solution: because 90° + 90° = 180°, as it satisfies the condition of supplementary angles. Two angles making a linear pair are always adjacent angles. Since, AB || l and EF is a transversal. An angle adjacent to a right angle is also a right angle. Two angles in a linear pair are adjacent to each other. Solution: True Since, 90° + 90° = 180°, a supplementary angle. Solution: ∴ ∠TRU + ∠QUR = 180° [Co-interior angles] (c) 70°, 110° ∴ x + 64° + 46° +100° – 360° (c) 30°, 50° In the given figure, l, m and n are parallel lines, and the lines p and q are also parallel. Solution: a and b are on the opposite side of transversal l. (a) interior angles on the same side of the transversal ∴ b = 50° [Alternate interior angles]. (a) 126° ⇒ a = 180° – 65° = 48° [Using (i)] Answered By . Two Adjacent Angle Can Be Complementary Too If They Add Up To 90°. Now, AB || CD and ED is a transversal. ∠AOB ∠ A O B + … ∴ ∠1 = ∠3 = 30° ——— (ii) [Corresponding angles] (b) 30° Solution: (d) adjacent but not supplementary [ ∵∠AOE = 30° and ∠DOB =40° (given)] ∴ ∠AOD + ∠AOC = 180° [Linear pair] Two right angles are complementary to each other. Hence, ∠a = 30°, ∠b = 150°, ∠c = 150°. In each of the following figures, write, if any, ∠1 +∠2 = 180° [Angles on a straight line PQ] (iii) uncommon arms are always opposite rays. Question 20. $$\Rightarrow \quad x=\frac{176^{\circ}}{4}=44^{\circ}$$ ⇒ 4x = 90° – 2 = 88° True, Question 69. (iv) ∠AOC and ∠AOD; ∠BOC and ∠BOD; ∠AOC and ∠BOC, ∠AOD and ∠BOD are adjacent angles. ∠EPQ+ ∠GQP = 130° + 50° = 180° ​, 18. (d) Since PQ I RS and line 1 is a transversal. Solution: (b) ∠d=∠c (a) complementary ⇒ 2x = 166° False (b) If one of the angles is obtuse, then other angle of a linear pair is acute. $$\Rightarrow x=\frac{180^{\circ}}{5}=36^{\circ}$$ a: If two lines intersect, then the vertically opposite angles are equal. ∴∠CHE = ∠HCB – 120° ———- (i) [Alternate interior angles] The 2 angles concerned don’t necessarily have to be adjacent, where the angles share a common point/vertex and a common side between them. $$\Rightarrow b=\frac{100^{\circ}}{5}=20^{\circ}$$. ∴ ∠COF = ∠EOD = 110° [Using (i)] [Vertically opposite angles] Now, ∠BOC = (x + 5)° = (35 + 5) = 40° True Solution: ⇒ d = 180°- 38° = 142° [Using (i)] Solution: ∴∠PAB = ∠ABC [Alternate interior angles] Thus, one angle is 100° and other is 80°. Obtuse, Question 52. ∠ABC is the complement of ∠CBD Supplementary Angles. Solution: 7 Answers. ⇒ x = 180° – 61° = 119° No, but they're "supplementary".Two "complementary" angles add up to 90Â°. toppr. Also, PQ || RS and line I is a transversal. Putting the value of x in (i), we get (d) Since, sum of the angles about a point is 360° ∴ ∠APR = ∠PRD [Alternate interior angles] In the given figure ∠AOB and ∠BOC are complementary as ∠AOB + ∠BOC = 30° + 60° = 90°. As per the Congruent supplement Theorem, two angles which are are supplementary to the same angles then those two sets of angles are congruent to each other. Solution: ⇒ ∠1 = 70° [Using (ii)] ∠2 = ∠4 [Corresponding angles] We have, find them. Question 83. Two lines AB and CD intersect at O (see figure). Solution: In the given figure, PA || BC || DT and AB || DC. ⇒ x+y – 90° ——- (i) Here we say that the two angles complement each other. (a) 36° In the given figure, PQ is a mirror, AB is the incident ray and BC is the reflected ray. If ∠POR = 90° and x : y = 3 : 2, then z is equal to (iii) There is no pair of vertically opposite angles and no angles are in the form of linear pair. My answer-- Vertical angles (e) Since, POQ is a straight line. (d) ∠a = ∠b [Vertically opposite angles] …, ar is her work place from herhouse?bor house​. (b) Since, PA || BC and AB is a transversal. ⇒ ∠BCD = 180°- 50° = 130° False ∠6 = ∠7 [Alternate exterior angles] Yes, as sum of two right angles is 1 8 0 o. (b): Since, vertically opposite angles are equal. A transversal intersects two or more than two lines at _________ points. Then, ∠QOS measures As, PQ is a straight line. 1 decade ago. $$\Rightarrow \quad a=\frac{180}{5}=36$$ Therefore, B will be less than 45°. ∴ ∠HCB =∠CDE [Corresponding angles] 100° + y = 180 ⇒ y = 180° – 100° = 80° Linear, Question 51. ∴ ∠PQR – ∠QRS = 85° ———– (i) [Alternate interior angles] Directions: In questions 57 to 71, state whether the statements are True or False. Find the angles. Solution: (b) 144° ⇒ ∠RSC = 180° – 65° = 115° ⇒ x = 180° – 66° = 114° ∴ x + y = 90° ———(i)     [Angles are complementary] and x – y = 30° ——— (ii) [Given] ⇒ 60° + ∠2 = 180° [Using (ii)] (b) p is true and q is false Name; Solution: Solution: As angles ∠QRS and ∠CSR are alternate interior angles and are equal. 40-50 20 (c) 122° ⇒ 5a – 20° = 180° Which of the following is false? When the sum of two angles is 180∘ 180 ∘, the angles are called supplementary angles. (d) ∠2 + ∠3 = 180° ∴ ∠ABC + ∠CDE = 180° [Co-interior angles] (c) ∠PQT and ∠TQS; ∠TQS and SQR; ∠PQT and ∠TOR; ∠PQS and ∠SQR are four pairs of adjacent angles. (ii) No, a and b are not adjacent angles as they don’t have common arm. $$\Rightarrow x=\frac{180^{\circ}}{3}=60^{\circ}$$ In the given figure, AB||CD. (c) 150° Now, EF || GH and AB is a transversal. Can two acute angles form a pair of supplementary angles? Since, l || m and p is a transversal. Question 8. According to question, ∠POR and ∠ROQ; ∠ROQ and ∠OOS; ∠QOS and ∠SOP; ∠SOP and ∠POR; ∠ROT and ∠TOS; ∠OOT and ∠POT are linear pairs. (a) pair of complementary angles We have ⇒ ∠2 = 180° -60° = 120° Each angle is called the supplement of the other angle. Measures (in degrees) of two complementary angles are two consecutive even integers. (a) 95°, 85° They are lined up in a straight line in front of the goalpost [See fig.(i)). You can sign in to vote the answer. Explain your answer. Since, PQ || RT and PR is a transversal. Thus the required angles are 90° each. ∴ ∠BCD = ∠CDT [Alternate interior angles] (b) Let the angle be x. (a) 60°, 120° Hence, ∠x = 35° and ∠y = 145°. (d) 105°, 75° (c) p is false and q is true For example: This is always true! Let A and B are two angles making a complementary angle pair and A is greater than 45° Question 61. ⇒ 60° – ∠L ——— (i) Find the values of a, b and c. Question 29. (c) 85° ∴ ∠2 + 135° = 180° [Co-interior angles] (d) making a linear pair Solution: Now, PO is a straight line. ∠a + ∠d = 180° [Linear Pair] The proposition continues by stating that on a transversal of two parallel lines, corresponding angles are congruent and the interior angles on the … (a) 110° Solution: (c) 5° (d) both p and q are false ∴ a = 36 $$\Rightarrow x=\frac{120^{\circ}}{2}=60^{\circ}$$ (A) 4000 km(B) 2400 km(C) 3000 km(D) 2700 km(E) None of theseGive me a proper explanation. ∴ a = 3x = 3 × 36° = 108° They form a right-angle triangle and making an L shape. (a) 95° Solution: Two angles are equal and su... maths. ⇒ 4 × 30° = 3b      [using (i)] Solution: They just need to add up to 90 degrees. (a) ∠1 + ∠5 = 180° For example, the player has a better chance of scoring a goal from Position A than from Position B. ⇒ (6x – 30) = 180 ⇒ 6x – 180 + 30 = 210 ∴ ∠b + ∠1 = 180° [Linear pair] So, two right angles are always supplementary to each other. (c) PQ || RS, line l is a transversal. ⇒∠APQ + ∠QPR = 130° Those two adjacent angles will always add to 180°. . Strictly sticking to traditional definitions, supplementary and complementary only refers to two angles… (c) 16° ⇒ 2∠ABP = 180° – 46° = 134 In figure, OB is perpendicular to OA and ∠BOC = 49°. No, two acute angles cannot form a pair of supplementary angles. Adding (i) and (ii), we get and 2x + 3 = 2 × 44° + 3 = 88° + 3 = 91° (c) 100°, 60° As if both angles are 89° and 89°, even then they cannot make the sum 180°. Now, CH || DF and CD is a transversal. (c) 45° ⇒ ∠POR + ∠QOS = 180° – 90° = 90° ——- (i) Three angles or more angles whose sum is equal to 90 degrees cannot also be called complementary angles. In the given figure, state which pair of lines are parallel. (d) ∠4 = ∠8 Both angles of a pair of supplementary angles can never be acute angles. ⇒ x + x = 180° [∵Angles are supplementary] (d) Since, PQ || RS, line l is a transversal. (d) 60° Solution: Question 42. ⇒ ∠2 = 180° – 75° = 105° ⇒ 2x + x = 180° Since, QP || RS and QR is a transversal. ∴ Let a = 3x and b – 2x The difference of two complementary angles is 30°. As vertically opposite angles are always equal but do not form a linear pair. Vertical angles always have equal measure. Solution: ⇒ 50° + ∠BCD = 180° (i) b and c (a) 20°, 50° Question 76. In the given figure, POR is a line. ∴ ∠COD = 90° Supplementary angles are two angles whose measures sum to 180 degrees. Solution: ⇒ ∠Q = 180° – 60° = 120°, Question 13. Then, sum of two right angles will be (90°+ 90°) = 180°. ∴ ∠POR + ∠ROT + ∠TOQ = 180° According to question, Two angles making a linear pair are always supplementary. These angles are on the same side of transversal CD. ⇒ ∠2 = 180° – (2a+ b)° ——– (ii) [∵ ∠1 = (2a + b)° (given)] This implies that there are interior angles on the same side of the transversal which are less than two right angles, contradicting the fifth postulate. Also, m || n and p is transversal. Vertical angles are across the intersection point from each other. ∠a = ∠3 [Vertically opposite angles] Thus, x = 110° and y = 100°. ∵ AB is a straight line. The angles x – 10° and 190° – x are the whole is greater than the sum of its parts. Question 3. As two right angles are supplementary to each other. Two lines intersect to form vertical angles. Here we have given NCERT Exemplar Class 7 Maths Solutions Chapter 5 Lines and Angles. False Parallel, Question 50. ∠1 ∠ 1 and ∠2 ∠ 2 are complementary if ∠1 +∠2 = 90∘ ∠ 1 + ∠ 2 = 90 ∘ Find the sum 2a +b. Two adjacent angles always form a linear pair. If ∠ABC = 46°, then ∠ABP is equal to ∠POR + 5 ∠POR = 90° ∠FOR + ∠QRH = 123° + 57° = 180° These two angles (140° and 40°) are Supplementary Angles, because they add up to 180°: Notice that together they make a straight angle. ∴ AOB is a straight line. (b) 11° Solution: (b) 46° Examples of complementary angles: 40° and 50° 60° and 30° 85° and 5° 70° and 20° 45° and 45° etc. Solution: ⇒ 720° = 5x Solution: ∴ ∠ABP + ∠ABC + ∠CBQ = 180° ⇒ x + 210° = 360° true. (d) 144° In the given figure, OR ⊥ OP. Solution: (d) 120° ⇒ 3x + 2x = 180° Solution: (ii) ∠x and ∠y are complementary angles. ∴ a = 20° [Alternate interior angles] ∠2 = ∠3 [Alternate interior angles] 8. ∠AOF=∠COD [Vertically opposite angles] Its complementary angle will be _________ ok i'm in GT math. The angle which makes a linear pair with an angle of 61° is of (b) ∠2 = ∠4 In a pair of complementary angles, each angle cannot be more than _________ (a) supplementary ⇒ 120° + ∠z= 180° [Using (i)] Solution: (d) (ii) is false (a) 125° (a) how many angles are formed? (c) Estimate atleast two situations such that the angles formed by different positions of two players are complement to each other. Now, CD and EF intersect each other at O. ⇒ ∠2 = ∠y = 120° [Vertically opposite angles] The angles between North and West and South and East are Question 19. ∴ ∠2 + 75° = 180° [Co-interior angles] Take any two adjacent angles from among the four angles created by two intersecting lines. True. Then, which of the following is true? Two adjacent angels are both acute angles. False. Now, p || q and n is a transversal. Complementary Angles: Much like supplementary angles, complementary angles add up to 90 degrees. Solution: Question 80. ⇒ ∠2 = 30° ———– (ii) ⇒ 42° + ∠QUR = 180° [Using (i)] (d) 120° q: a and b are forming a pair of adjacent angles. We have, Solution: True. (d) 45°, 35° ∴ 6y + y + 2y = 180° ∴ ∠POR + ∠ROS + ∠QOS = 180° (iii) d and f The greater the angle, the better chance the player has of scoring a goal. Let one angle be 2x and the other angle be 2x + 2. (a) 4th player has the greatest kicking angle. Question 57. ∴ Let each angle be x. $$\Rightarrow a=\frac{200^{\circ}}{5}=40^{\circ}$$, Question 14. (i) ∠AOD and ∠DOB; ∠DOB and ∠BOC, ∠BOC and ∠AOC; ∠AOC and ∠AOD are four pairs of supplementary angles. Now, LM || QP and QT is a transversal. (d) In figure (d), a and bare adjacent angles since, they have a common vertex, common arm and also, their non-common arms are on either side of common arm. Question 85. (c) Since, AB || CD and PR is a transversal They cannot. A. ∴ Its supplement = 180° – x RD Sharma Class 11 Solutions Free PDF Download, NCERT Solutions for Class 12 Computer Science (Python), NCERT Solutions for Class 12 Computer Science (C++), NCERT Solutions for Class 12 Business Studies, NCERT Solutions for Class 12 Micro Economics, NCERT Solutions for Class 12 Macro Economics, NCERT Solutions for Class 12 Entrepreneurship, NCERT Solutions for Class 12 Political Science, NCERT Solutions for Class 11 Computer Science (Python), NCERT Solutions for Class 11 Business Studies, NCERT Solutions for Class 11 Entrepreneurship, NCERT Solutions for Class 11 Political Science, NCERT Solutions for Class 11 Indian Economic Development, NCERT Solutions for Class 10 Social Science, NCERT Solutions For Class 10 Hindi Sanchayan, NCERT Solutions For Class 10 Hindi Sparsh, NCERT Solutions For Class 10 Hindi Kshitiz, NCERT Solutions For Class 10 Hindi Kritika, NCERT Solutions for Class 10 Foundation of Information Technology, NCERT Solutions for Class 9 Social Science, NCERT Solutions for Class 9 Foundation of IT, PS Verma and VK Agarwal Biology Class 9 Solutions, NCERT Solutions for Class 10 Science Chapter 1, NCERT Solutions for Class 10 Science Chapter 2, Periodic Classification of Elements Class 10, NCERT Solutions for Class 10 Science Chapter 7, NCERT Solutions for Class 10 Science Chapter 8, NCERT Solutions for Class 10 Science Chapter 9, NCERT Solutions for Class 10 Science Chapter 10, NCERT Solutions for Class 10 Science Chapter 11, NCERT Solutions for Class 10 Science Chapter 12, NCERT Solutions for Class 10 Science Chapter 13, NCERT Solutions for Class 10 Science Chapter 14, NCERT Solutions for Class 10 Science Chapter 15, NCERT Solutions for Class 10 Science Chapter 16, CBSE Previous Year Question Papers Class 12, CBSE Previous Year Question Papers Class 10. In the given figure, AB || CD, AF || ED, ∠AFC = 68°and ∠FED = 42°. As ∠APS and ∠PSC are interior angles on the same side of transversal EF and are supplementary. Question 79. Two right angles are always supplementary to each other. In A Right Triangle, The Altitude From A Right-angled Vertex Will Split The Right Angle Into Two Adjacent Angle; 30°+60°, 40°+50°, Etc. Then Find ∠ABC and ∠CDE. ⇒ 100° + a = 180° sum of interior angles on the same side of a transversal is _________ ∠1 and ∠8; ∠2 and ∠7; ∠3 and ∠4; ∠4 and ∠5; ∠5 and ∠6; ∠3 and ∠6 are six pairs of supplementary angles. Thus, both the angles are of 83°. ⇒ ∠COA = 90° – 49° [∵ ∠BOC = 49° (given)] (a) If one of the angles is acute, then other angle of a linear pair is obtuse. (i)∠1 = 65° [Vertically opposite angles] If the two complementary angles are adjacent then they will form a right angle. Solution: Solution: These angles are opposite to each other and always equal. corresponding angles are on the _________ side of the transversal. The value of a is A linear pair may have two acute angles. $$\Rightarrow \quad x=\frac{180^{\circ}}{4}=45^{\circ}$$ (d) 62° Now, PQ || RT and RQ is a transversal. The supplementary angles. When two lines cross, they form two pairs of vertical angles. $$\Rightarrow \quad b=\frac{120^{\circ}}{3}=40^{\circ}$$ In the given figure, PO || RS. (c) four pairs of adjacent angles. (b) ∠2 + ∠5 =180° Solution: If ∠1 = (2a + b)° and ∠6 (3a – b)°, then the measure of ∠2 in terms of b is Since, l || m and q is a transversal. and its supplement = 180° – x= 180°- 100 = 80° (b) 24° (ii) each linear pair. Question 59. (iv) Let the angle between c and fig ∠4. (b) 45° (d) 80° ⇒ y = 180° – 48° = 132° Question 104. Now, a + b = 45° + 55° = 100°, Question 105. Find the value of a + b. Question 33. (iii) Two right angles cannot be complement of each other. ∠ ABD is a compliment of ∠ DBC because; ∠ ABD + ∠ DBC =90° (right angle). ∴ ∠2 = 30° [Corresponding angles] In the given figure, if l || m, find the values of a and b. ⇒ ∠CDE-120° ——- (ii) [Using (1)] Question 78. Three lines AB, CD and EF intersect each other at O. ⇒ 3x = 180° – x ⇒ 3x + x = 180° sometimes. If one of them is one-third of the other, find the angles. NCERT Exemplar Class 7 Maths Chapter 5 Lines and Angles are part of NCERT Exemplar Class 7 Maths. In the given figure, if QP || SR, the value of a is ∴ y + 48° = 180° [Co-interior angles] Yes, 2 right angles can form a supplementary angle. Question 96. Also, AB || CD and EF is a transversal. ⇒ 85° + a = 180° [Using (ii)] Now, SP || RQ and PR is a transversal. but ∠4 ≠ ∠8, Question 38. ⇒2x = 180° – x the legs of an isosceles triangle are equal. Question 73. if equal quantities are multiplied by equal quantities the products are ____ eual. Since, l || m and q is a transversal ∴ ∠c + ∠2 = 180° [Linear pair] In the given figure, l || m || n. ∠OPS = 35° and ∠QRT = 55°. Let one angle be 2x + 1, then the other angle is 2x + 3. (i) ∠1 and ∠3; ∠2 and ∠4; ∠5 and ∠7; ∠6 and ∠8 are four pairs of vertically opposite angles. Solution: As one acute angle and one obtuse angle can make two supplementary angles. (a) Let the two angles be x and 2x. We have, ⇒ ∠QPR = 130° – 50° = 80°, Question 11. If due toheavy traffic the average speed of the bus is20 km/h, how f Solution: Now, ∠2-(3a – b)° [From (i)] (ii) one arm is always common, and Solution: Question 21. (d) 30° (a) 35° ⇒ x = 360°- 210° Solution: ∴ x + 66° = 180° [Co-interior angles] ∴ ∠1 = ∠3 [Corresponding angles] (c) 80° Also, m is a straight line. ∴ b + 132° = 180° [Co-interior angles] ∴ ∠RST + a = 180° [Linear pair] ⇒ a = 180° – 85° = 95°. Angles between South and West and South and East are ∴ EF and GH are not parallel lines. ⇒ x + 2x = 300° Also, AB || DF and BD is a transversal. Solution: (d) vertically opposite angles Yes. (b) a is true and b is false ⇒ 5b – 180° – 80° = 100° (iv) No, a and b are not adjacent angles as the arms which are not common are on the same side of common arm. ∴ x + 2y + 3y = 180° ⇒ 5a + 130° = 180° ∴ b = 48° ——— (i) ⇒ 6 ∠POR = 90° (i) ∠x and ∠y; ∠x and ∠y + ∠z; ∠y and ∠z; ∠z and ∠x + ∠y are four pairs of adjacent angles. (i) Let the angle between b and c is ∠1. Thus, the required angle is 11°. (d) 119° (c) write all the pairs of vertically opposite angles. Since, a transversal intersects two parallel lines, then interior angles on the same side of a transversal are supplementary. Statements a and bare as given below: $$\Rightarrow c=\frac{120^{\circ}}{4}=30^{\circ}$$ ———– (i) (a) ∠1 = ∠3 sometimes. $$\Rightarrow \quad x=\frac{88^{\circ}}{4}=22^{\circ}$$ Can two angles be supplementary, if both of them be right? If an angle is 60° less than two times of its supplement, then the greater angle is - 32510872 (i) EF and GH ∴ ∠4 = 75° [Alternate interior angles], Question 91. (c) 136° Now, SOT is a straight line (ii) There is no pair of vertically opposite angles. Thus, ∠BOC = 40°, Question 101. ∴ ∠PQR + ∠RQO = 180° [Linear pair] ⇒ ∠c = 180° – 30° = 150° (c) 90° Now, RS is a straight line. (b) how many types of angles are formed? (a) both p and q are true Complementary angles always have positive measures. (d) Vertically opposite angles are always equal. S. Two vertical angles are congruent. 45° : Given, angle = 45° Thus, the greater angle is 100°. (b) 15° Then, angles a and bare respectively Also, RQ || TS and RS is a transversal. Hence, in this case these two angles are considered congruent. 2x + 1 + 2x + 3 = 180° Question 111. ⇒ ∠2 = (3 × 36 – b)° = (108 – b)°, Question 36. Let the angle be x. Two right angles are always supplementary to each other. Madhu leaves her house at 9 a.m. and reachesher workplace by bus at 10:30 a.m. ∴ Its complement = 90° – x (d) supplementary (a) one of its angles is acute? Adding (i) and (ii), we get As ∠RSP and ∠QPD are corresponding angles and are not equal. always. In the given figure, ∠ROS is a right angle and ∠POR and ∠QOS are in the ratio 1 : 5. ⇒ 30° + 5y = 180° If two supplementary angles are in the ratio 1 : 2, then the bigger angle is Solution: As interior angles on the same side of a transversal with two distinct parallel lines are supplementary angles. ∠ABD and ∠DBC; ∠ABE and ∠CBE are linear pairs. An angle is more than 45°. ∴ a = 65° [Alternate interior angles] ∴ These angles are complementary. (b) 108° false . ∴ ∠POQ + ∠QOR = 180° [Linear pair] (d) 150° Question 22. Then If you have any query regarding NCERT Exemplar Class 7 Maths Solutions Chapter 5 Lines and Angles, drop a comment below and we will get back to you at the earliest. ∴ (∠2+42°) + ∠3 = 180° [Co-interior angles] (d) 80°,100 Question 112. Question 66. Now, l || m and AC is a transversal. 1 decade ago. Solution: alternate interior angles have one common _________