# two right angles can make a pair of complementary angles

∴ 90°- x = 79° (b) 144° Now, l || m and n is a transversal. Measures (in degrees) of two supplementary angles are consecutive odd integers. Now, e || f and d is a transversal. ∠d = ∠c [Vertically opposite angles] (c) 64° (c) Draw a line L.M passing through T such that LM || QP || SR. Now, CH || DF and CD is a transversal. (a) 95° ⇒ 4 × 30° = 3b      [using (i)] ∴ a = 65° [Alternate interior angles] Question 7. Then, angles a and bare respectively As ∠RSP and ∠QPD are corresponding angles and are not equal. Question 4. A + B = 90° (b) (3 – 6)° (c) (108 – b)° ⇒ b = 180° – 132° = 48° (d) 101° The S in supplementary can be used to form the 8 in 180. We have, (a) Since, ∠P + ∠Q = 180° (d) ∠2 + ∠3 = 180° (b) PQ is a straight line. Solution : False Measure of right angle is 90°. (b) ∠d=∠c When talking about complementary angles, it's important to remember that they're always in apair. $$\Rightarrow a=\frac{50^{\circ}}{5}=10^{\circ}$$, Question 27. corresponding angles are on the _________ side of the transversal. A pair of complementary angles is angles . ∴ a = 45° [Corresponding angles] ⇒ x = 360°- 210° The difference of two complementary angles is 30°. (d) 135° ∴ b = 70° [Corresponding angles] ⇒ b = c – a ∴ x + 64° + 46° +100° – 360° Question 80. True, Question 64. We have, 3k + 2k = 90° ⇒ 5k = 90° (iii) ∠1 and ∠2, ∠3 and ∠4, ∠5 and ∠6 are three pairs of supplementary angles. We have, A right angle measures 90°. ⇒ ∠2 = 180° – 42° – 68° [Using (i)] ⇒ 4x = 180° Now, 2a + b = 2 × 132° + 132° 2x + 1 + 2x + 3 = 180° ⇒ ∠ABC = 180° – 120° [Using (ii)] x + 61° = 180° [Linear pair] Also, m is a straight line. (b) 50°,130° (ii) Name all the pairs of complementary angles. Thus, the required angles are 60° and 30°. ⇒ 3x = 180° ∴ ∠1 + ∠5 = 180° [By(i)and(iii)] So an angle of 30° has a supplementary angle of 90° - 30° = 60°. $$\Rightarrow \quad a=\frac{120^{\circ}}{6}=20^{\circ}$$ Thus, a = 67° and b = 48°, Question 102. ⇒ ∠1 = 70° [Using (ii)] ⇒ 3a – b + 2a + b = 180 ⇒ 5a = 180 Complementary Angles. ⇒ ∠2 = 180° -60° = 120° $$\Rightarrow x=\frac{180^{\circ}}{3}=60^{\circ}$$ Now, a || d and c is a transversal. ∴ b = 50° [Alternate interior angles]. Then, And that is because their measures add up to 90 degrees. ⇒ x + x = 180° [∵Angles are supplementary] (c) p is false and q is true If ∠AOE = 30° and ∠DOB = 40° (see figure), find ∠COF. 180°, Question 46. ⇒ 5b – 180° – 80° = 100° (b) supplementary $$\Rightarrow x=\frac{166^{\circ}}{2}=83^{\circ}$$ Complementary, Question 43. We have As, PQ is a straight line. Solution: If the sum of measures of two angles is 180° then they are _________ (d) 120° 4.Two angles that are right are always congruent. ⇒ x + 2x = 300° ⇒ 85° + a = 180° [Using (ii)] (b) 100° You can check whether an angle is a right angle by using a set square, like in the picture below. Since, l || m and q is a transversal (b) 67° Solution: because 90° + 90° = 180°, as it satisfies the condition of supplementary angles. ∴ Angles between South and West and South and East are making a linear pair. (c) Since, angles are on a straight line. ⇒ 50° + ∠BCD = 180° Take any right angle and draw in a ray that has its endpoint as the vertex of the right angle. ∴ EF || GH, Question 113. In the given figure, if QP || SR, the value of a is (a) ∠1 + ∠5 = 180° ∴∠PAB = ∠ABC [Alternate interior angles] sum of interior angles on the same side of a transversal is _________ Thus, the angle which is half of its supplement is of 60°. Solution: (ii) Let the angle between d and e is ∠2. (b) 50°, 20° Since, vertically opposite angles are equal. If the sum of two angles are 90 o then the angles are said to be Complementary angles. (d) 20° In the given figure, ∠ROS is a right angle and ∠POR and ∠QOS are in the ratio 1 : 5. Solution: ∴ x + 2x = 180° (i) ∠AOD and ∠DOB; ∠DOB and ∠BOC, ∠BOC and ∠AOC; ∠AOC and ∠AOD are four pairs of supplementary angles. They can be adjacent angles but don’t have to be. (a) 13 angles are formed. We know that the sum of the measures of the supplementary angles is 180°. ∴ y + 48° = 180° [Co-interior angles] Answer. (a) vertically opposite angles (d) equal (b) 61° ADC and BDC are . Since, 90° + 90° = 180°, a supplementary angle. ⇒∠APQ + ∠QPR = 130° (b) (iii) is false (b) complementary Solution: (a) 120° Also, AB || CD and EF is a transversal. Two angles making a linear pair are always supplementary. Solution: An angle which is half of its supplement is of ∴ ⇒ 50° + ∠QPR – 130° Question 77. ∴ a = 36 (a) ∠TOS and ∠SQR is a pair of complementary angles. Two right angles are complementary to each other. Solution: ∴ ∠TRU + ∠QUR = 180° [Co-interior angles] ∴ b = 48° ——— (i) Solution: 82° _____ 5. ∴ ∠POQ + ∠QOR = 180° [Linear pair] Thus, one angle is 44° and other is 46°. As one acute angle and one obtuse angle can make two supplementary angles. Question 70. but ∠a ≠ ∠d, Question 12. If the sum of two angles is 180 degrees then they are said to be supplementary angles, which forms a linear angle together.Whereas if the sum of two angles is 90 degrees, then they are said to be complementary angles, and they form a right angle together. (i) Yes, and b are the adjacent angles as they have a common vertex, one common arm and other non-common arms on the opposite side of the common arm. Also, PQ || RS and line I is a transversal. Now, l || m and AC is a transversal. Question 73. (e) Since, POQ is a straight line. Solution: (b) 50° If two angles add up to 90°, they are _____ angles. (c) 36° Statements a and bare as given below: ∴ 2x + 1 = 2 × 44° + 1 = 88° + 1 = 890 (b) Now the players are lined up as shown in Fig. (c) alternate interior angles ⇒ 3x = 300° There is an easy way to try and remember these using the first letters of each word. Opposite, Question 49. alternate interior angles have one common _________ Two angles forming a _________ pair are supplementary. ⇒ 5 ∠POR = ∠QOS ——– (ii) (d) 22.5° An angle is 45°. Solution: (b) how many types of angles are formed? Two adjacent angles always form a linear pair. (c) 60° (c) 72° Two Angles are Supplementary when they add up to 180 degrees. Also, AB || DF and BD is a transversal. Now, ∠2-(3a – b)° [From (i)] ∴ ∠3=68° ——– (i) [Corresponding angles] ∴ These angles are complementary. Solution: (a) 120° In the given figure, line I intersects two parallel lines PQ and RS. ∴ ∠ABP = ∠CBQ ——– (1) ⇒ ∠c + 30° = 180° [Using (iii)] Solution: ⇒ 90° + x + y = 180° False In the given figure, P, Q and R are collinear points and TQ ⊥ PR, Solution: (d) (ii) is false A supplementary angle usually involves two angles that are adjacent and adds up to 180 degrees. In the given figure, if AB || CD, ∠APQ = 50° and ∠PRD = 130, then ∠QPR is (d) 119° Question 19. Solution: (a) 90° Thus, a = 50° and b = 130°. Give reason in support of your answer. ⇒ x = 90° – 79° ⇒ x – 11° (b) If one of the angles is obtuse, then other angle of a linear pair is acute. The legs of a stool make an angle of 35″ with the floor as shown in figure. (b) 90°, 90° Question 42. (i) vertex is always common, Proof Write the correct one. ∠1 = ∠2 ——– (iii) [Verticallyopposite angles] Now, 60° + ∠1 = 60° + 120° = 180° and these angles are interior angles on the same side of transversal l. ⇒ ∠2 = 180° – (2a+ b)° ——– (ii) [∵ ∠1 = (2a + b)° (given)] (i) ∠1 and ∠3; ∠2 and ∠4; ∠5 and ∠7; ∠6 and ∠8 are four pairs of vertically opposite angles. ∴ ∠2 = ∠6= (34 – b)° ——– (i)[Corresponding angles] The supplement of the right angle is always _________ angle. Linear, Question 51. Solution: x = 2(180° – x) – 60° Question 21. The greater the angle, the better chance the player has of scoring a goal. (a) Let an angle be x. Solution: Then, the angles are (c) 55° In the given figure, PQ||RS and a : b = 3 : 2. Also, ∠EFG = ∠1 + ∠2 – 34° +45° = 79° ⇒ 3x = 180° – x ⇒ 3x + x = 180° ⇒ z + 36° = 180° ⇒ z = 180° – 36° = 144°, Question 31. ⇒ a = 180° – 100° = 80°. (b) We know that angle of incident and angle of reflection is same. (d) Since, sum of the angles about a point is 360° (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. True, Question 63. (ii) each linear pair. In figure, OB is perpendicular to OA and ∠BOC = 49°. ∴ ∠ABP + ∠ABC + ∠CBQ = 180° (d) ∠a = ∠b [Vertically opposite angles] l || m and PQ is a transversal janaeaholloway janaeaholloway 05/06/2019 Mathematics Middle School Use the diagram to identify the special angle pairs. (d) ∠3 = ∠7 Since, QP || RS and QR is a transversal. The value of bis Solution: Thus, a = 20°, b = 40° and c = 30°, Question 109. ∴ ∠QRT = ∠RQU [Alternate interior angles] Solution: In the given figure, if PQ || RS and QR || TS, then the value of a is Putting value of x in (i), we get and its supplement = 180° – x= 180°- 100 = 80° Supplementary angles: Two angles that add up to 180° (or a straight angle) are supplementary. Question 25. ⇒ ∠c = 180° – 30° = 150° What are the measures of the angle and its supplement? (b) ∠2 = ∠4 Since, two angles form a linear pair. In the given figure, find the value of ∠BOC, if points A, O and B are collinear. (d) 154° Solution: Solution: (c) one of its angles is right? Determine the values of x and y. If one of them is one-third of the other, find the angles. In the given figure, PQ is a mirror, AB is the incident ray and BC is the reflected ray. Find ∠PQR. ∴ ∠AOD = 139°, Question 94. True. Find the measures of two complementary angles if one angle is five times the other angle. ∴ ∠4 = 75° [Alternate interior angles], Question 91. Therefore, B will be less than 45°. ⇒ 60° + 20 = 180° ∴ Supplement of x = 180° – 28° = 152°, Question 88. $$\Rightarrow \angle A B P=\frac{134^{\circ}}{2}=67^{\circ}$$. $$\Rightarrow \quad y=\frac{180^{\circ}}{9}=20^{\circ}$$, Question 17. 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