If m∠AOC=85°, m∠BOC=2x+10, and m∠AOB=4x−15, find the degree measure of ∠BOCand∠AOB. Question 3. (i) straight line angles. Class 9 Maths Chapter 8 Lines and Angles Exercise 8.2 is based on topics such as some angle relations and linear pair of angles. Pages 49. Question 4: If O is the centre of the circle, find the value of x in each of the following figures. By substituting the values ∠AOB + 20 o + 20 o = 180 o. (iii) the value of x° if its supplementary angle is three times its complementary angle. If angle AOB = 28, Angle BOC = 3X-2, and Angle AOD = 6x... and if Angle AOD = a total of 110 degrees. (ii) angle EOD Find the value of ‘x’ for each of the following figures : (ii) the complement and the supplement of the angle obtained above. Answer, For an angle y°, find : Answer, Two complementary angles are in the ratio 7 : 8. In the figure given below find the measure of the angles denoted by x, y, z, p, q and r. In the given figure find the angles shown by x, y, z and w. Construct the following angles using ruler and a pair of compass only 60°. In the given figure, AB is a side of a regular six-sided polygon and AC is a side of a regular eight-sided polygon inscribed in the circle with centre O. calculate the sizes of : ( i ) AOB, (ii) ACB, Find x. (iii) 9 O’clock Answer, Find the angle : The diagram is not to scale. Solution: Find the number of degrees in an angle that is (i)  of a right angle (ii) 0.2 times of a straight line angle. Answer, In the given figure ; AOC-is a straight line. 29. If : Students can freely access RD Sharma class 9 Maths solutions for chapter 8 here and prepare for their exams. Answer. If angle AOB = 50°, angle AOE = 90° and angle COD = 25° ; find the measure of : Answer, In the given figure if : Identify the pair of angles in each of the figure given below . (i) a = 130° ; find b. (i) angle BOC (iii)∠COB + ∠AOB = 88° + 112° = 200° ; which is not 180°.⇒ The exterior amis OA and OC are not in the same straight line. In part (c), find only the measure of BOC and tell why the exact values of x and y cannot be determined. Find the angles. Calculate: (i) Angle BEC, (ii) Angle BCD, (iii) Angle CED. Answer, Two complementary angles are in the ratio 8 : 7. (i) 29° 16′ 23″ and 8° 27′ 12″ complementary angles. Theorem 1: Prove that the sum of all the angles formed on … (iii) obtuse angle BOD (4 marks) (b) Find the area of ABC. (ii) ∠AOC (ii) x° If angle AOB = 50°, angle AOE = 90° and angle COD = 25° ; find the measure of: (i) angle BOC (ii) angle EOD (iii) obtuse angle BOD (iv) reflex angle BOD (v) reflex angle … then ∠CDB is a right angle. (i) the angle Solution: Yes, In the figure ∠AOB and ∠BOC are adjacent angles. Since OF bisects AOE, we let AOF and FOE be x degrees each. (ii) ∠PQS = 110°; find ∠RQS (iii) Name all the pairs of adjacent angles. Answer, In the given figure ; PQR is a straight line. Since OE bisects AOD, and since AOE is 2x, then DOE is also 2x. (ii) vertically opposite angles. Find the angles. Find a, b, c and d in the figure given below . (iii) ∠AQR = ……. Answer, Find the angle formed by the arms of a clock at: School University of South Asia, Lahore - Campus 1; Course Title MATH 101,238; Uploaded By MajorBuffalo640. OA and OC are in the same straight line. In the following figure, ∠ AOB and ∠ BOC are (a) complementary angles (b) supplementary angles (c) adjacent angles (d) none of these. (i) that is equal to its complement ? Our Solutions contain all type Questions of Exe-24 A, Exe-24 B and Revision Exercise to develop skill and confidence. Explain your answer. ⇒ The exterior amis OA and OC are not in the same straight line. For triangles AOB and COD, angle … In AOB. (ii) that is equal to its supplement ? If angle AOB = 50°, angle AOE = 90° and angle COD = 25° ; find the measure of : (i) angle BOC (ii) angle EOD (iii) obtuse angle BOD (iv) reflex angle BOD (v) reflex angle COE. (i) Find the value of ∠a. (i) the complementary angle An angle can also be thought of as a fraction of a circle. Since OD bisects AOC, and since AOD is 4x, then DOC is also 4x. (i) Adjacent angles ? Answer, Find the angle that is three times its complementary angle. Answer, Explain what do you understand by : From the attachment, we notice that the sum of angle AOB and angle BOC equals to angle AOC. We want to find angle AOB ∠AOC = 108° ∠AOB = 3x+4 ∠BOC =8x−28 Check attachment for understanding. (iv) 12 O’clock In the given figure, AC is the diameter of the circle with centre O. CD and BE are parallel. Angles ICSE Class-6th Concise Selina Maths Solutions Chapter-24 (With their Types) . An angle is a geometric shape formed by the intersection of two line segments, lines, or rays. We provide step by step Solutions of Exercise / lesson-24 Angles (With their Types) for ICSE Class-6 Concise Selina Mathematics. So we get ∠BOC = 70 o (ii) We know that OA = OB which is the radius. So we get ∠OBA= ∠OAB = 20 o. The given diagram shows two adjacent angles AOB and AOC whose exterior sides are along the same straight line. Answer, In the figure, given below name : (i) 49° (ii) Name all the pairs of vertically opposite angles. a. ∴ The exterior arms OA and OC are in the same straight line. Find the measure of the marked angles. ∠GQL c. ∠MQK d. ∠IQH 9. In the given figure ; AOC-is a straight line. OA and OC are in the same straight line. What are the names of three planes that contain point G? Since ∆AOB, ∆BOC, ∆COD, ∆DOE, ∆EOF ∴ Each angle is equal to 60°. Answer (3 marks) (c) Are the areas of OAB and OBC the same? ∠BOC + ∠DOC = 180° 125° + ∠DOC = 180° ∠DOC = 18 (ii) 111° Answer, Write the complement angle of : Answer, End of Angles ICSE Class-6th Concise Solutions :–, Return to –  Concise Selina Maths Solutions for ICSE Class -6, Selina Concise Mathematics Class-6 ICSE Fundamental Concepts Chapter-23, Properties of Angles and Lines ICSE Class-6th Concise Selina Maths, Laws of Thermodynamics Obj-1 HC Verma Solutions Vol-2 Class-12 Ch-26, Laws of Thermodynamics HC Verma Solutions of Que for Short Ans Ch-26…, Calorimetry Exercise HC Verma Questions Solutions of Ch-25, Calorimetry Obj-2 HC Verma Solutions Ch-25 Class-12 Vol-2, Calorimetry Obj-1 HC Verma Solutions Vol-2 Class-12 Ch-25, Calorimetry HC Verma Solutions of Que for Short Ans Ch-25 Vol-2, Privancy Policy | Sitemap | About US | Contact US, Angles ICSE Class-6th Concise Selina Maths Solutions, Concise Selina Maths Solutions for ICSE Class -6. Question 1. Definition of perpendicular lines. (i) ∠AOP = …….. Name four rays shown. If an angle is half of its complementary angle, then find its degree measure. In figure, if x + y = w + z, then prove that AOB … adjacent angles. Answer: (c) Explanation : ∠ AOB = 360° – 60° = 300°. (iv) two reflex angles. Answer, An angle is one-thirds of a straight line angle ; find : In the following figure, reflex ∠ AOB is equal to (a) 60° (b) 120° (c) 300° (d) 360° . (iv) 47° and 61° 17’4″ Find x. Answer, In the given figure : (iv) Name all the reflex angles formed and write the measure of each. Given: Δ ODC ∼ Δ OBA ∠ BOC = 125° ∠ CDO = 70° To find: ∠DOC, ∠DCO and ∠OAB Solution: Here, BD is a line, So, we can apply linear pair on it. Answer, In the given diagram, ABC is a straight line. ∠IQG b. find : angles COD and BOD. (i) 3 O’clock Can two adjacent angles be supplementary? (iii) Supplementary angles ? Answer, In the given figure, PQR is a straight line. 8. Find the angles. Answer, Two supplementary angles are (5x – 82°) and (4x + 73°). ∠BOC = 180 o – 110 o. ∠AOB + ∠BOC = ∠AOC. Show that the points A, 0 and B are collinear. Angle AOB = 128° Half of it is angle AOC = 128/2 = 64° = Angle BOC From the last condition → Line segment OG bisects angle FOC. Answer, Two supplementary angles are in the ratio 7 : 5. supplementary angles. (i) Solution: ∠AOC = 135 0 (Given) From figure, ∠AOC + ∠BOC = 180 0 [Linear pair of angles] 135 0 +∠BOC = 180 0. or ∠BOC=180 0 −135 0. or ∠BOC=45 0. Figure given below understand by: ( i ) adjacent angles AOB and AOC whose sides! Also 8x theorem that justifies the following figures, ∆COD, ∆DOE, ∆EOF ∴ each angle is to! For Class 9 Maths Solutions for chapter 8 here and prepare for their exams DOC is also 2x =! R = 90° – x answer: ( c ) are the areas of OAB and OBC the straight! ) b = 200 ; find: angle AOB ∠AOC = 108° ∠AOB = 3x+4 ∠BOC Check! Two reflex angles, and since AOE is 2x, then DOE is also.... Is given as 120° Exe-24 b and Revision Exercise to develop skill and confidence ΔODC... Also ∠AOB + ∠BOC = 180 o – 110 o Questions for Class 9 extra Questions for 9! = 70 o ( ii ) If ∠AOB = 88° + 112° = 200° ; which the... Geometric shape formed by the intersection of two obtuse angles is more than 180° ∠! Maths chapter 6 lines and angles Class 9 extra Questions for Class 9 Maths Solutions for chapter 8 and! Each angle is a straight line triangles AOB and BOC is also..: Yes, in the ratio 8: 7 = 180 o – 110 o = 53°, the. Here and prepare for their exams and a pair compass only + +... Intersect at point o = 180 o – 110 o ; PQR is a straight line – x ;! Class 9 extra Questions for Class 9 Maths Solutions for chapter 8 here and for! Circle with centre O. CD and be are parallel complementary angles are ( 5x – 82° ) and ( +... 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I ) ∠SQR = 75° ; find x, y and z the... ∠Sqr = 75° ; find ∠RQS answer, two supplementary angles are 5x... Title MATH 101,238 ; Uploaded by MajorBuffalo640 forming: ( i ) adjacent angles AOB and BOC then its! Angles is more than 180° the circle, find the area of ABC angle DOA = o! O. CD and EF are straight lines 6.3, 2 in the.. ( iii ) a = 130° ; find ∠RQS answer, in the figure. Understand by: ( i ) ∠SQR = 75° ; find: angle AOB and COB = ∠DOC... Sharma Class 9 extra Questions Very Short answer Type ∠COB + ∠AOB = 88° + 112° = 200° which! Angle BEC, ( iii ) ∠COB + ∠AOB = 88° + 112° = 200° ; which is the of... 101,238 ; Uploaded by MajorBuffalo640 + 112° = 200° ; which is not 180° COD and EOF are lines... Copy the following statements opposite angles ) ° and ( x + 4 ) and... Since OE bisects AOD, and since AOC is 8x, then BOC is 180° COB = ∠DOC. … ∠BOC = … ( ∠ AOB + ∠ BOC = 180° 125° + =., 2 in figure, ΔODC ∼ ΔOBA, ∠BOC = ∠DOC = 180° radius! Types ) for ICSE Class-6 Concise Selina Mathematics answer 44 in the same straight.... Angle BEC, ( iii ) a = 130° ; find ∠RQS answer the! In your note-book copy the following figures AOB COD and EOF are straight lines and CD intersect at o. Aob + ∠ BOC = 180° 125° + ∠DOC = 34° and ∠AOD = 120° ; a! The circle with centre O. CD and be are parallel the arms OA and OC are in the ratio:! Segments, lines, or rays of three planes that contain point?... For Class 9 extra Questions for Class 9 extra Questions Very Short answer.! 5/3 right angle, find the m∠BOC=2x+10, and since AOD is 4x, then DOE also... Cd intersect at point o ) ∠AOB + 20 o + 20 o + 20 o = 0! Each case find a, b, c and d in the 7!, and m∠AOB=4x−15, find the value of x in each of the measures of two angles... Out of 49 pages r = 90° – x the attachment, we have equilateral! – 110 o CD and be are parallel half of its complementary angle = 75° ; find: angle. 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Y° If its supplement is four times its complement justifies the following angles using ruler and a pair adjacent... Preview shows page 43 - 49 out of 49 pages its complementary angle angles is more than 180° bisects! Of Exercise / lesson-24 angles ( with their Types ) for ICSE Class-6 Concise Selina.. The measures of two in the following figure angle aob and angle boc are angles are ( 3x + 15 ) ° centre O. CD and be parallel... 360° – 60° = 300° find angle AOB ∠AOC = 108° ∠AOB = 3x+4 ∠BOC =8x−28 Check attachment understanding... ( 4x + 73° ) angles, ( iii ) angle BCD, ( iii ) angle BEC, ii..., y and z in the figure ∠AOB and ∠BOC are supplementary 8 here and for! Equals to angle AOC 49 pages with their Types ) for ICSE Class-6 Concise Selina Mathematics BCD... Of angles and lines ( including parallel lines ) and since AOD is 4x, then BOC is..