Selina Concise Class 6 Math Chapter 23 Properties Of Angles And Lines Exercise 23A Solution
EXERCISE 23A
(1) Two straight lines AB and CD intersect each other at a point O and angle AOC = 50o; find:
(i) Angle BOD = 50o; because ∠AOC and ∠BOD are opposite angle.
(ii) ∠AOD = 180o – ∠BOD = (180 – 50)o = 130o
(iii) ∠BOC = 130o; because ∠BOC and ∠AOD are opposite angle.
(2) The adjoining figure shows two straight lines AB and CD intersecting at point P. If ∠BPC = 4x – 5o and ∠APD = 3x + 15o, find:
(i) 4x – 5 = 3x + 15 (∵∠CPB = ∠APD; opposite angles)
⇒ 4x – 3x = 15 + 5
⇒ x = 20
(ii) ∠APD = 3x + 15
= (3 × 20) + 15
= 60 + 15 = 75o
(iii) ∠BPD = 180 – ∠BPC
= 180 – (4x – 5)
= 180 – 4x + 5
= 185 – (4 × 20)
= 185 – 80 = 105o
(iv) ∠BPC = 4x – 5
= (4 × 20) – 5
= 80 – 5
= 75o
(3) The figure, given alongside, shows two adjacent angles AOB and AOC whose exterior sides are along the same straight line. Find the value x.
Solution: 68o + 3x – 20o = 180o
⇒ 3x + 48o = 180o
⇒ 3x = 180o – 48o
⇒ 3x = 132o
⇒ x = 44
(4) Each figure, given below, shows a pair of adjacent angles AOB and BOC. Find whether or not the exterior arms OA and OC are in the same straight line.
(i) 90 – x + 90 + x = 180
⇒ 180 = 180
Yes
(ii) 97 + 83 = 180
⇒ 180 = 180
Yes
(iii) 68 + 112 = 180
⇒ 180 = 180
Yes
(5) 5x – 40 + x + 10 = 180
⇒ 6x – 30 = 180
⇒ 6x = 180 + 30
⇒ 6x = 210
⇒ x = 35
Therefore, ∠APB
= 5x – 40o
= (5 × 35)o – 40o
= 175o – 40o
= 135o