# SSC – Lines and Angles Syllabus Notes 2021: Download SSC – Lines and Angles Syllabus Study Materials

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### Solved Example

1. In the adjacent figure lines AB and CD intersects at O. If ∠AOC + ∠BOE = 70° and ∠BOD = 40° then find∠BOE and reflexive ∠COE

Solution: ∠AOC = ∠BOD (Vertically opposite angle)

∴ ∠AOC = 40°

According to question, ∠AOC + ∠BOE = 70° or,

40° + ∠BOE = 70°

∴∠BOE = 30°

Now, ∠COD + ∠DOB + ∠BOE = reflexive ∠COE

∴180° + 40° + 30° = reflexive∠COE

∴reflexive ∠COE = 250°

1. In the figure given below lines XY and MN intersect at O.

If LPOY = 90° and a : b = 2 : 3 then find the measure of c.

Solution: Given,

Let a = 2k and b = 3k

∵∠POY = 90°                   ∴ ∠POX = 90°

or, ∠a + ∠b = 90°              or, 2k + 3k = 90°

or, 5k = 90°                       or, = 18°

∴∠b = 3k = 3 × 18° = 54°

Now, ZXOM = LYON (Vertically opposite angle)

∴∠YON = 54°

Again, ∠XON + ∠YON = 180° (Linear pair of angles axiom)

or, c + 54° = 180°              ∴c = 126°

1. Given that ∠XYZ = 64° and line XY is produced to point P. Draw a diagram from the given information. If ray YQ bisects ∠ Then find the measure of ∠XYQ and reflex ∠QYP.

Solution: ∠XYZ + ∠ZYQ + ∠QYP = 180° (Linear pair of angles)

or, 64° + 2∠ZYQ = 180° (∵∠XYZ = 64° and ∠ZYQ = ∠QYP)

or, 2∠ZYQ = 116°

∴∠ZYQ = 58°;

∵ ∠XYQ = ∠XYZ + ∠ZYQ

∴∠XYQ = 64° + 58° = 122°

Now reflex, ∠QYP = ∠PYX + ∠XYQ

= 180° + 122° = 302°

1. In the figure given below find x and y and hence prove that AB||CD.

Solution:∠DRS = ∠CRQ (Vertically opposite angle)

∴y = 130°

Also, ∠ASP + ∠ASR = 180°

(Linear pair of angles)

or, 50° + x = 180°

or, x = 130°

∵x = y = 130°

AB||CD (Alternate angle)

1. In the given figure if PQ||ST, ∠PQR = 110° and ∠RST = 130°, Find ∠QRS

Solution: From point R draw a line RM parallel ST.

∠RST + ∠SRM = 180° 1 (Consecutive interior angle)

or, 130° + ∠SRM = 180° ∴∠SRM = 50° … (i)

Now, ∠QRM = ∠PQR

or, ∠QRM = 110°             or, ∠QRS + ∠SRM = 110°

or, ∠QRS + 50° = 110° (from (i)) ∴∠QRS = 60°

1. In the given figure if AB||CD, ∠APQ = 50° and ∠PRD = 127°, find x and y.

Solution: AB||CD             ∴ LAPR = LPRD (Alternate angle)

or, ∠APQ + ∠QPR = ∠PRD

or, 50° + y = 127° or, y = 77°

and ∠APQ = ∠PQR         (Alternate angle)

or, 50° = x, ∴ x = 50°

1. In ∠ABC, ∠A = 40°. If bisector of AB and AC meets at O then prove that ∠BOC = 110°

Solution: In∠ABC,

∠A + ∠B + ∠C = 180°

or, 40° + ∠B + ∠C = 180°

or, ∠B + ∠C = 140°

or,                                       … (i)

Now, In ∆BOC, ∠OBC + ∠OCB + ∠BOC = 180°

or,  + ∠BOC = 180°or, 70° + ∠BOC = 180°                                           (from (i))

∠BOC = 110° Proved.

[Shortcut: BOC = 90° += 90° + 20° = 110° Proved.]

1. If angles of a triangle are is the ratio 2:3:4, then find the least and greatest angle.

Solution: Let angle be 2x°, 3x° and 4x°.

∴2x° + 3x° + 4x° = 180°

or, 9x° = 180°                    or, x° = = 20°

∴least angle = 2x° = 2 × 20° = 40°

greatest angle = 4x° = 4 × 20° = 80°

1. The exterior angle of a triangle is 110° and one of its interior opposite angle is 30°, find other angles.

Solution: Consider the triangle ABC in which exterior ∠ACD = 110° and ∠A = 30°, we have to find ∠B and ∠C.

∵∠ABC + ∠BAC = ∠ACD

or, ∠ABC + 30° = 110°

or, ∠ABC = 80°

or, ∠ACB + ∠ACD = 180°

or, ∠ACB + 110° = 180°

∵ ∠ACB = 70°

1. In triangle PQR, sides QP and RQ respectively produced to point S and T. If ∠SPR = 135° and ∠PQT = 110° find ∠PRQ

Solution:∠SPR + ∠QPR = 180° (linear pair of angles)

or, 135° + ∠QPR = 180°

or, ∠QPR = 45°                                        … (i)

Now, ∠QPR + ∠PRQ = ∠PQT

or, 45° + ∠PRQ = 110° (from(i))

∴∠PRQ = 65°

1. In the adjacent figure ∠X = 62° and ∠XYZ = 54°. If VO and ZO respectively bisects ∠XYZ and ∠XZY then find ∠OZY and ∠YOZ

Solution: In ∆XYZ, ∠YXZ + ∠XYZ + ∠XZY = 180°

or, 62° + 54° + ∠XZY = 180°

∴ ∠XZY = 64°

From question,

∠OYZ = ∠XYZ = × 54° = 27°

and ∠OZY = ∠XZY = × 64° = 32°

Now, In ∆OYZ,

∠OYZ + ∠OZY + ∠YOZ = 180°

or, 27° + 32° + ∠YOZ = 180°

∴∠YOZ = 121°

1. In the given figure lines PQ and RS intersect at point T such that ∠PRT = 40°; ∠RPT = 95° and ∠TSQ = 75° Find ∠SOT

Solution: In ∆PRT, ∠RPT + ∠PRT = ∠PTS

or, 95° + 40° = ∠PTS

∠PTS = 135°                                             … (i)

Again in,

∆TSQ, ∠TSQ + ∠SQT = ∠PTS

or, 75° + ∠SQT = 135° (∵∠TSQ = 75° and ∵∠PTS = 135°)

∠SQT = 135° – 75° = 60°.

1. In the figure given below if PQ⊥PS, PQ||SR, ∠SQR = 28° and ∠QRT = 65° find x and y.

Solution:∵PQ||SR

∠PQR = ∠QRT (Alternate angle)

or, x + 28° = 65°

or, x = 65°-28° = 37°                                 … (i)

Now, In

∆PQS, ∠SPQ + ∠PQS + ∠QSP = 180°

or, 90° + x + y = 180°

or, 90° + 37° + y = 180°                (from (i))

or, y = 180° -127° = 53°

Hence, x = 37° and y = 53°.