📝 Important Formulas, Rules & Key Points

• Angle Sum Property of a Triangle: The sum of all interior angles of a triangle is always 180°.
  i.e., In ΔABC, ∠A + ∠B + ∠C = 180°
• Exterior Angle Theorem: If a side of a triangle is produced, then the exterior angle so formed is equal to the sum of the two interior opposite angles.
  i.e., Ext. ∠ACD = ∠A + ∠B
• Linear Pair Axiom: If a ray stands on a line, then the sum of two adjacent angles so formed is 180°.

Question 1

In the given figure, sides QP and RQ of ΔPQR are produced to points S and T respectively. If ∠SPR = 135° and ∠PQT = 110°, find ∠PRQ.

Solution:

1. Since TQR is a straight line, the angles ∠PQT and ∠PQR form a linear pair.
   ⇒ ∠PQT + ∠PQR = 180°
   ⇒ 110° + ∠PQR = 180°
   ⇒ ∠PQR = 70°
2. Similarly, SPR is a straight line, so ∠SPR and ∠QPR form a linear pair.
   ⇒ ∠SPR + ∠QPR = 180°
   ⇒ 135° + ∠QPR = 180°
   ⇒ ∠QPR = 45°
3. Now, in ΔPQR, by the Angle Sum Property:
   ⇒ ∠QPR + ∠PQR + ∠PRQ = 180°
   ⇒ 45° + 70° + ∠PRQ = 180°
   ⇒ 115° + ∠PRQ = 180°
   ⇒ ∠PRQ = 180° - 115°
   ⇒ ∠PRQ = 65°

Answer: ∠PRQ = 65°

Question 2

In the given figure, ∠X = 62°, ∠XYZ = 54°. If YO and ZO are the bisectors of ∠XYZ and ∠XZY respectively of ΔXYZ, find ∠OZY and ∠YOZ.

Solution:

1. In ΔXYZ, by the Angle Sum Property:
   ⇒ ∠X + ∠XYZ + ∠XZY = 180°
   ⇒ 62° + 54° + ∠XZY = 180°
   ⇒ 116° + ∠XZY = 180°
   ⇒ ∠XZY = 180° - 116° = 64°
2. Since ZO is the bisector of ∠XZY:
   ⇒ ∠OZY = ½ × ∠XZY = ½ × 64° = 32°
3. Since YO is the bisector of ∠XYZ:
   ⇒ ∠OYZ = ½ × ∠XYZ = ½ × 54° = 27°
4. Now, in ΔYOZ, by the Angle Sum Property:
   ⇒ ∠OYZ + ∠OZY + ∠YOZ = 180°
   ⇒ 27° + 32° + ∠YOZ = 180°
   ⇒ 59° + ∠YOZ = 180°
   ⇒ ∠YOZ = 180° - 59° = 121°

Answer: ∠OZY = 32° and ∠YOZ = 121°

Question 3

In the given figure, if AB || DE, ∠BAC = 35° and ∠CDE = 53°, find ∠DCE.

Solution:

1. Since AB || DE and AE acts as a transversal line:
   ⇒ ∠DEC = ∠BAC (Alternate Interior Angles)
   ⇒ ∠DEC = 35°
2. Now, in ΔCDE, by the Angle Sum Property:
   ⇒ ∠DCE + ∠CDE + ∠DEC = 180°
   ⇒ ∠DCE + 53° + 35° = 180°
   ⇒ ∠DCE + 88° = 180°
   ⇒ ∠DCE = 180° - 88° = 92°

Answer: ∠DCE = 92°

Question 4

In the given figure, if lines PQ and RS intersect at point T, such that ∠PRT = 40°, ∠RPT = 95° and ∠TSQ = 75°, find ∠SQT.

Solution:

1. In ΔPRT, by the Angle Sum Property:
   ⇒ ∠PRT + ∠RPT + ∠PTR = 180°
   ⇒ 40° + 95° + ∠PTR = 180°
   ⇒ 135° + ∠PTR = 180°
   ⇒ ∠PTR = 180° - 135° = 45°
2. Since lines PQ and RS intersect at T, the vertically opposite angles are equal:
   ⇒ ∠STQ = ∠PTR = 45°
3. Now, in ΔTSQ, by the Angle Sum Property:
   ⇒ ∠TSQ + ∠STQ + ∠SQT = 180°
   ⇒ 75° + 45° + ∠SQT = 180°
   ⇒ 120° + ∠SQT = 180°
   ⇒ ∠SQT = 180° - 120° = 60°

Answer: ∠SQT = 60°

Question 5

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

Solution:

1. Since PQ || SR and QR is the transversal line:
   ⇒ ∠PQR = ∠QRT (Alternate Interior Angles)
   ⇒ x + 28° = 65°
   ⇒ x = 65° - 28° = 37°
2. Since PQ ⊥ PS, ∠SPQ = 90°.
   In right-angled triangle ΔSPQ, by the Angle Sum Property:
   ⇒ ∠SPQ + x + y = 180°
   ⇒ 90° + 37° + y = 180°
   ⇒ 127° + y = 180°
   ⇒ y = 180° - 127° = 53°

Answer: x = 37° and y = 53°

Question 6

In the given figure, the side QR of ΔPQR is produced to a point S. If the bisectors of ∠PQR and ∠PRS meet at point T, then prove that ∠QTR = ½∠QPR.

Proof:

1. For ΔPQR, ∠PRS is an exterior angle.
   By Exterior Angle Theorem:
   ⇒ ∠PRS = ∠QPR + ∠PQR
   Multiplying both sides by ½:
   ⇒ ½∠PRS = ½∠QPR + ½∠PQR   — (Equation 1)
2. Since QT is the bisector of ∠PQR, ½∠PQR = ∠TQR.
   Since RT is the bisector of ∠PRS, ½∠PRS = ∠TRS.
   Substituting these in Equation 1:
   ⇒ ∠TRS = ½∠QPR + ∠TQR   — (Equation 2)
3. Now, for ΔTQR, ∠TRS is an exterior angle.
   By Exterior Angle Theorem:
   ⇒ ∠TRS = ∠QTR + ∠TQR   — (Equation 3)
4. From Equation 2 and Equation 3, we equate the values of ∠TRS:
   ⇒ ∠QTR + ∠TQR = ½∠QPR + ∠TQR
   Subtracting ∠TQR from both sides:
   ⇒ ∠QTR = ½∠QPR

Hence Proved!

🚀 Aptitude Corner (For NTSE & Olympiads)

When solving geometry problems in competitive exams, you can save valuable time using these standard direct results:

• Angle Bisector Theorem (Internal): If OB and OC are internal angle bisectors of ∠B and ∠C in ΔABC, then:
  ∠BOC = 90° + ½∠A
• Angle Bisector Theorem (External): If OB and OC are external angle bisectors of ∠B and ∠C in ΔABC, then:
  ∠BOC = 90° - ½∠A
• Mixed Bisector Relation (Question 6 standard): The angle made by the internal bisector of one base angle and the external bisector of the other base angle is always half of the vertical angle.

Congratulations! More solutions are waiting for you to discover.

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