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- 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°
- 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.
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°
- 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°
- 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)
- 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°
- 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°
- In ∠ABC, ∠A = 40°. If bisector of AB and AC meets at O then prove that ∠BOC = 110°
∠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.]
- 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°
- 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°
- 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°
- 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°
∠OYZ = ∠XYZ = × 54° = 27°
and ∠OZY = ∠XZY = × 64° = 32°
Now, In ∆OYZ,
∠OYZ + ∠OZY + ∠YOZ = 180°
or, 27° + 32° + ∠YOZ = 180°
∴∠YOZ = 121°
- 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)
∆TSQ, ∠TSQ + ∠SQT = ∠PTS
or, 75° + ∠SQT = 135° (∵∠TSQ = 75° and ∵∠PTS = 135°)
∠SQT = 135° – 75° = 60°.
- In the figure given below if PQ⊥PS, PQ||SR, ∠SQR = 28° and ∠QRT = 65° find x and y.
∠PQR = ∠QRT (Alternate angle)
or, x + 28° = 65°
or, x = 65°-28° = 37° … (i)
∆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°.