### Lines And Angles Class 9th Mathematics AP Board Solution

##### Question 1.In the given figure three lines and intersecting at O. Find the values of x, y and z it is being given that x : y : z = 2 : 3 : 5 Answer:From the given, the three angles are x,y,zIf the two lines intersect at a point then its vertically opposite angles are equal.∴ A = x, B = z and C = yWe know that,The sum of all the angles around at a point is equal to 360°∴ A + B + C + x + y + z = 360°⇒ x + y + z + x + y + z = 360°⇒ 2x + 2y + 2z = 360°⇒ 2(x + y + z) = 360°⇒ (x + y + z) = ⇒ x + y + z = 180° ------(1)Given that, x : y : z = 2 : 3 : 5Let x = 2m,y = 3m,z = 5m (∵ m = constant)Substitute these values in equation (1) we get2m + 3m + 5m = 18010m = 180m = ∴ m = 18Substituting m = 18 in x,y,zx = 2m,x = 2(18) = 36°y = 3m,y = 3(18) = 54°z = 5m,z = 5(18) = 90°∴ x = 36°,y = 54°,z = 90°Question 2.Find the value of x in the following figures.i. ii. iii. iv. Answer:(i) From the given figure,3x + 18° + 93° = 180°⇒ 3x + 111° = 180°⇒ 3x = 180°-111°⇒ 3x = 69°⇒ x = ∴ x = 24°(ii) From the given figure,(x-24)° + 29° + 296° = 360°⇒ (x-24)° = 360°-325°⇒ (x-24)° = 35°⇒ x = 35° + 24°∴ x = 59°(iii) From the given figure,(2 + 3x)° = 62°⇒ 3x = 62°-2° = 60°⇒ x = ∴ x = 20°(iv) From the given figure,40° + (6x + 2)° = 90°⇒ (6x + 2)° = 90°-40°⇒ (6x + 2)° = 50°⇒ 6x = 50°-2° = 48°⇒ x = ∴ x = 8°Question 3.In the given figure lines and intersect at O. If ∠AOC + ∠BOE = 70° and ∠BOD = 40°, find ∠BOE and reflex ∠COE. Answer:Given that,The lines and intersect at O.∠AOC + ∠BOE = 70° ----(1)∠BOD = 40° ----(2)If the two lines intersect at a point then its vertically opposite angles are equal.∴ ∠AOC = ∠BODSubstitute (2) in (1)⇒ 40° + ∠BOE = 70°⇒∠BOE = 70°-40°∴∠BOE = 30°From the figure,AOB is a straight line and its angle is 180°So, ∠AOC + ∠BOE + ∠COE = 180°From equation (1)⇒ 70° + ∠COE = 180°⇒∠COE = 180°-70°∴∠COE = 110°Reflex ∠COE = 360° - 110° = 250°∴∠BOE = 30° and Reflex ∠COE = 250°Question 4.In the given figure lines and intersect at O. If ∠POY = 90° and a: b = 2 : 3, find c. Answer:Given that,The lines and intersect at O.From the figure, XOY is a straight line and its angle is 180°So, ∠XOM + ∠MOP + ∠POY = 180°From the given, Let ∠a = 2x and ∠b = 3x⇒∠b + ∠a + ∠POY = 180°----(1)Given that ∠POY = 90°Substitute the values in equation (1),⇒2x + 3x + 90° = 180°⇒5x + 90° = 180°⇒5x = 180°-90° = 90°⇒x = ∴ x = 18°⇒ ∠a = 2x = 2×18° = 36°⇒ ∠b = 3x = 3×18° = 54°From the figure, MON is a straight line and its angle is 180°⇒∠b + ∠c = 180°⇒54° + ∠c = 180°⇒∠c = 180°-54°∴ ∠c = 126°Question 5.In the given figure ∠PQR = ∠PRQ, then prove that ∠PQS = ∠PRT. Answer:In the figure, ST is a straight line and its angle is 180°So, ∠PQS + ∠PQR = 180° ----(1)And ∠PRT + ∠PRQ = 180°-----(2)From the two equations, we get∠PQS + ∠PQR = ∠PRT + ∠PRQGiven that,∠PQR = ∠PRQ⇒ ∠PQS + ∠PRQ = ∠PRT + ∠PRQ⇒ ∠PQS = ∠PRT + ∠PRQ -∠PRQ⇒ ∠PQS = ∠PRTSo, ∠PQS = ∠PRT is proved.Question 6.In the given figure, if x + y = w + z, then prove that AOB is a line. Answer:In a circle, the sum of all angles is 360°∴ ∠AOC + ∠BOC + ∠DOB + ∠AOD = 360°⇒ x + y + w + z = 360°Given that, x + y = w + z⇒ w + z + w + z = 360°⇒ 2w + 2z = 360°⇒ 2(w + z) = 360°⇒ w + z = 180° or ∠DOB + ∠AOD = 180°If the sum of two adjacent angles is 180° then it forms a line.So AOB is a line.Question 7.In the given figure is a line. Ray is perpendicular to line is another ray lying between rays and  Prove that Answer:Given that, is a line and is perpendicular to line .⇒ ∠POR = 90°The sum of linear pair is always equal to 180°∴ ∠POS + ∠ROS + ∠POR = 180°Substitute ∠POR = 90°⇒90° + ∠POS + ∠ROS = 180°⇒∠POS + ∠ROS = 90°∴ ∠ROS = 90°-∠POS----(1)⇒ ∠QOR = 90°Given that OS is another ray lying between OP and OR⇒∠QOS-∠ROS = 90°∴∠ROS = ∠QOS-90°----(2)On adding two equations (1) and (2) we get2∠ROS = ∠QOS-∠POS⇒ ∠ROS = (∠QOS-∠POS)So, ∠ROS = (∠QOS-∠POS) is proved.Question 8.It is given that ∠XYZ = 64° and XY is produced to point P. A ray YQ bisects ∠ZYP. Draw a figure from the given information. Find ∠XYQ and reflex ∠QYP.Answer:Let us draw a figure from the given, Given that, a ray YQ bisects ∠ZYPSo, ∠QYP = ∠ZYQHere, PX is a straight line, so the sum of the angles is equal to 180°∠XYZ + ∠ZYQ + ∠QYP = 180°Given that, ∠XYZ = 64° and ∠QYP = ∠ZYQ⇒ 64° + 2∠QYP = 180°⇒ 2∠QYP = 180°-64° = 116°∴ ∠QYP = = 58°Also, ∠QYP = ∠ZYQ = 58°Using the angle of reflection,∠QYP = 360°-58° = 302°∠XYQ = ∠XYZ + ∠ZYQ⇒∠XYQ = 64° + 58° = 122°∴ Reflex ∠QYP = 302° and ∠XYQ = 122°

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