Given:            O is the centre of the circle. ∡APB and ∡AQB are two inscribed angles subtended by the same arc AB. To prove:          ∡APB = ∡AQB. Construction: Join AO and OB .....

Given:            O is the centre of the circle. ∡QOR and ∡QRP are central angle and inscribed angle             subtended by the same arc QR. To Prove:       .....

Given:            O is the centre of the circle. ∡AOB and ∡ACB are central angle and inscribed angle subtended by the same arc AB. To Prove:            ∡AOB = 2∡ACB. Construction: .....

Here, two circles with radii of different measurements are drawn. In every figure, ∡BAC and ∡BDC are angles at the circumference in the same segment BC. The angles are measured with protractor and shown in the given .....

Here, two circles with radii of different measurements are drawn. In every figure, ∡BAC and ∡BDC are angles at the circumference of a circle standing on the same arc BC. The angles are measured with protractor and shown in the given .....

Here, two circles with radii of different measurements are drawn. In every figure, ∡BAC and ∡BOC are angles at the circumference and central angle of a circle standing on the same arc BC. The angles are measured with protractor and shown in .....

Here, two circles with radii of different measurements are drawn. In every figure, ∡BAC and ∡BOC are angles at the circumference and central angle of a circle standing on the same arc BC. The angles are measured with protractor and shown in .....

Here, two circles with radii of different measurements are drawn. In every figure, cyclic quadrilateral PQRS are made. ∡PQR and ∡PSR are opposite angles of cyclic quadrilateral PQRS. The angles are measured with protractor and shown in the .....

Here, two circles with radii of different measurements are drawn. In every figure, ∡ACB is the inscribed angle on the semi-circle. The angles are measured with protractor and shown in the given table: Table .....

Here, two circles with radii of different measurements are drawn. In every figure, a cyclic quadrilateral ABCD is constructed and CD is produced to E. ∡ADE is an exterior angle and ∡ABC is a interior angle of the cyclic quadrilateral .....

Here, two circles with radii of different measurements are drawn. In every figure, cyclic quadrilateral DEFG are made. The angles are measured with protractor and shown in the given table: Table : Figure ∡D ∡F ∡D + .....

Given:           O is the centre of the circle. ABCD is the cyclic quadrilateral. To prove:          ∡BAD + ∡DCB = 180° and ∡ABC + ∡ADC = 180° Construction: Join BO and DO .....

We have given,        ∡ROS = 120o. We know,          ∡SPR = 12⨉ ∡ROS                       .....

Here, ∡ADB = ∡ACB = 60° [Inscribed angles in the same arc] ∡DAC = 90° [Inscribed angle in the half circumference] In triangle, ABD, ∡ADB+∡DAC+∡DBA = 180°[Angles of triangle] or, 60°+90°+∡DBA = 180° or, ∡DBA = .....

Here,      ∡DAB+∡DCB = 180°[Opposite angle of cyclic quadrilateral] or, ∡DCB = 180°- 92°   ∴ ∡DCB = 88° ∡EBC  = ∡DCB = 88°[Alternate angles] ∡ADC = ∡CBF         .....

Here, ∡OTC = ∡TCO = 31°[In ∆OCT, OC=OT, radii of the circle] ∡OTP+∡OTC   = 180° [Linear pair] or, ∡OTP+31°   = 180° ∴ ∡OTP = 149° ∡OQP = 90°[Radius is perpendicular to tangent]   .....

Here, ∡PRQ = ∡PRS = 90o          [Angle suspended by diameter in circumference is equal] Now, If ⊥ of a vertex bisects the base line, then PQS is  an isoceleous triangle. So, ∡PRS = 60o      .....

We have given,        ∡ACB = 55o. We know,         ∡AOB = 2 ⨉ ∡ACB                             .....

Given,         ∡CDE = 80o           ∡CBE = ? We have,              ∡CBE + ∡CDE = 180o  [Opposite angles of a cyclic quadrilateral]   Or ,  .....

In triangle, OQR,      ∡QOR +∡OQR+∡ORQ = 180° [angles of triangle] or, ∡QOR+50°+50° = 180° [∆OQR is isosceles triangle] or, ∡QOR = 180°-100°                 .....

Given:                Here, in the given cyclic quadrilateral PQRS, QR is the bisector of ∡SQT.  To prove:            PR = SR (or ∆PSR is an isosceles .....

Figure (copy) Given:             PC is the bisector of ∡APB To prove:             XY//AB Construction: Join PQ.  Proof:   Statement Reason .....

Given:          AB = AC  To prove:          BE = DC . Proof:   Statement Reason .....

Here, ∡AOB = 2 × ∡CAB              = 2 × 55°             = 110° In triangle, OCB, ∡COB+∡OCB+∡OBC = 180° or, 110°+∡OCB+∡OBC = 180° or, .....

Given:           O is the   centre of circle.            DO⊥ AB. To prove:           ∡ AEC = ∡ODA .   Statement Reason .....

Given:          Here, in the cyclic quadrilateral ABCD, AC = BC . To prove:          DC is the bisector of   ∡ BDE. Proof:   Statement Reason .....

We have,  The pair of supplementary angles in the given figure are: ∡ADC and ∡ABC .          [Opposite angles of a cyclic quadrilateral] ∡DAB and ∡DCB.           [Opposite angles of .....

We have,        ∡ADB = 90o [A tangent to a circle is perpendicular to the radius of circle drawn at the point of contact] Now, In triangle, AOB;        x + y + ∡ADB = 180o [The sum of angles of triangle .....

Here, ∡ROQ+230° = 360° [Complete angle] ∴∡ROQ = 130° Again,     ∡ROQ+∡ORQ+∡OQR = 180° or, 130°+∡OQR+∡OQR = 180° or, 2∡OQR = 50° ∴   ∡OQR = 25°. ∡POQ = ∡OQR = .....

Here, ∡EDG = 12 × ∡EOG             = 12 × 140°             = 70° . Again,  ∡EDG+∡EFG = 180°.....i)[Opposite angles of cyclic .....

Let, ∡ABC=x and ∡AOC=y  So,      y = 2x . And, y = 360-x [Complete angle] Equating y,       2x = 360°-x or, 3x = 360° ∴    x = 120°, .....

Given, ∡AQB = 40o. We have,      ∡APB = ∡AQB  [Inscribed angles standing on the same arc] ∴  x = 40o. .....

Here, ∡DAO + 118° = 180° [Co-interior angle] ∴ ∡DAO = 62° ∡BAD = ∡ECD = 72° x + ∡DAO = ∡BAD or, x + 62° = 72° ∴   x = .....

Here, two circles with radii of different measurements are drawn. In every figure, cyclic quadrilateral DEFG are made. ∡D and ∡F are opposite angles and ∡E and ∡G are opposite angles. The angles are measured with protractor and shown in .....

Given,        ∡QOR = 100o       ∡ORP = 15o       ∡QRP = ? Now, In triangle OQR,        OQ=OR [Radii of same circle]        ∡QOR .....

Given,      ∡ABE =85o.     x = ? We have,          ∡ABE + ∡EBC = 180o. [The angles in straight line]         ∡EDC + ∡EBC = 180o.  [Opposite angles of a cyclic .....

We have,  ∡ADB = 90  [The angle in the circumference of a semi-circle is right angle] Now, In triangle, ABD x + y + ∡ADB = 180o [The angles of the triangle]             .....

Here,    BA = 2×OA           = 2×OC           = 2×5cm           = 10cm [ Diameter is twice the radius]     ∆BCA is .....

We have, The pair of equal angles in the given figure are:            ∡AQB and ∡APB                      [Inscribed angles .....

In triangle, BCD,     ∡CBD+∡BCD+∡CDB=180° [Angles of triangle] or, 33°+∡BCD+33°=180 ∴    ∡BCD = 180°-66°                   .....

Here, two circles with radii of different measurements are drawn. In every figure, cyclic quadrilateral ABCD are made. ∡A and ∡C are opposite angles of cyclic quadrilateral ABCD. The angles are measured with protractor and shown in the given .....

We have given,          ∡ACB = 60o.         ∡ADB = ∡ACB    [Inscribed angles standing on the same arc.]  So,        ∡ADB= 60o. .....

Here,      ∡ABC = ∡ADC = 50°[Co-interior angle]      ∡ABC+∡AEC = 180°[Opposite angles of cyclic quadrilateral] or, 50°+∡AEC = 180° ∴    ∡AEC = .....

Here, ∡ABD = a [Alternate angle] ∡ACD + ∡ABD = 180° [Opposite angle of cyclic quadrilateral] or, 130° + a = 180°  ∴   a = 50°. .....

Here, ∆PQS is right angled triangle  5x + 4x = 90°      [Two acute angles of the right angled triangle PQS] or, 9x = 90°     ∴ x = 10° ∡PRQ = ∡PSQ = 4x       [Inscribed .....

Here, ∡ABC = ∡AEB = ∡DCE = 72°    [Isosceles triangle and corresponding angles] ∡BAE+∡ ABC+∡AEB = 180°     [angles of triangle] or, ∡BAE+72°+72° = 180° ∴∡BAE = 36° In triangle .....

Here, ∡CBA = 2 × ∡ABD             = 2 × 43°             =86°[Given]. Again,      ∡BDA = ∡ACB = 67°[Inscribed angle on same arc]   .....

Here,         AC = b = 4 cm        AB = c = 7 cm        ∠B = 40o  and ∠C = 75o   So,∠A = 180o - ∠B -∠C               = .....

Let,        BC = distance of the car from the observer = 60 m        AB = height of the tower       ∠DAC = ∠ACB = 40o In right angled triangle ABC,      Tan40o = .....