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In figure, O is the centre of the circle. If ∠APB = 50°, find ∠AOB and ∠OAB.
∠APB = 50°
By degree measure theorem
∠AOB = 2∠APB
⇒ ∠APB = 2 × 50° = 100° since OA = OB [Radius of circle]
Then ∠OAB = ∠OBA [Angles opposite toequalsides]
Let ∠OAB = x
In ΔOAB, by angle sum property
∠OAB + ∠OBA + ∠AOB = 180°
⟹ x + x + 100° = 180°
⟹ 2x = 180° - 100°
⟹ 2x = 80°
⟹ x = 40°
∠OAB = ∠OBA = 40°
In figure, it is given that O is the centre of the circle and ∠AOC = 150°. Find ∠ABC.
∠AOC = 150°
∴ ∠AOC + reflex ∠AOC = 360° [Complex angle]
⇒ 150° + reflex ∠AOC = 360°
⇒ reflex ∠AOC = 360° − 150°
⇒ reflex ∠AOC = 210°
⇒ 2∠ABC = 210° [By degree measure theorem]
⇒ ∠ABC = 210°/2 = 105°
In figure, O is the centre of the circle. Find ∠BAC.
We have ∠AOB = 80°
And ∠AOC = 110°
Therefore, ∠AOB + ∠AOC + ∠BOC = 360° [Complete angle]
⇒ 80° + 100° + ∠BOC = 360°
⇒ ∠BOC = 360° − 80° − 110°
⇒ ∠BOC = 170°
∠BOC = 2∠BAC
⇒ 170° = 2∠BAC
⇒ ∠BAC = 170°/2 = 85°
If O is the centre of the circle, find the value of x in each of the following figures.
(i) ∠AOC = 135°
∴ ∠AOC + ∠BOC = 180° [Linear pair of angles]
⇒ 135° + ∠BOC = 180°
⇒ ∠BOC = 180° − 135°
⇒ ∠BOC = 45°
∠BOC = 2∠CPB
⇒ 45° = 2x
⇒ x = 45°/2 = 22½°
(ii) We have ∠ABC = 40°
∠ACB = 90° [Angle in semi circle]
In ΔABC, by angle sum property
∠CAB + ∠ACB + ∠ABC = 180°
⇒ ∠CAB + 90° + 40° = 180°
⇒ ∠CAB = 180° − 90° − 40°
⇒ ∠CAB = 50°
Now, ∠CDB = ∠CAB [Angle is same in segment]
⇒ x = 50°
(iii) We have
∠AOC = 120° By degree measure theorem.
∠AOC = 2∠APC
⇒ 120° = 2∠APC
⇒ ∠APC = 120°/2 = 60°
∠APC + ∠ABC = 180° [Opposite angles of cyclic quadrilaterals]
⇒ 60° + ∠ABC = 180°
⇒ ∠ABC = 180° − 60°
⇒ ∠ABC = 120°
∴ ∠ABC + ∠DBC = 180° [Linear pair of angles]
⇒ 120 + x = 180°
⇒ x = 180° − 120° = 60°
(iv) We have
∠CBD = 65°
∴ ∠ABC + ∠CBD = 180° [Linear pair of angles]
⇒ ∠ABC = 65° = 180°
⇒ ∠ABC = 180° − 65° = 115°
∴ reflex ∠AOC = 2∠ABC [By degree measure theorem]
⇒ x = 2 × 115°
⇒ x = 230°
(v) We have
∠OAB = 35° Then,
∠OBA = ∠OAB = 35° [Angles opposite to equal radii]
In ΔAOB, by angle sum property
⇒∠AOB + ∠OAB + ∠OBA = 180°
⇒ ∠AOB + 35° + 35° = 180°
⇒ ∠AOB = 180° −35° − 35° = 110°
∴ ∠AOB + reflex ∠AOB = 360° [Complexangle]
⇒ 110° + reflex ∠AOB = 360°
⇒ reflex ∠AOB = 360° − 110° = 250°
By degree measure theorem reflex
∠AOB = 2∠ACB
⇒ 250° = 2x
⇒ x = 250°/2 = 125°
(vi) We have
∠AOB = 60° By degree measure theorem reflex
⇒ 60° = 2∠ACB
⇒ ∠ACB = 60°/2 = 30° [Angles opposite to equal radii]
⇒ x = 30°.
(vii) We have
∠BAC = 50° and ∠DBC = 70°
∴ ∠BDC = ∠BAC = 50° [Angle in same segment]
In ΔBDC, by angle sum property
∠BDC + ∠BCD + ∠DBC = 180°
⇒ 50° + x + 70° = 180°
⇒ x = 180° − 50° − 70° = 60°
(viii) We have
∠DBO = 40° and ∠DBC = 90° [Angle in a semicircle]
⇒ ∠DBO + ∠OBC = 90°
⇒ 40° + ∠OBC = 90°
⇒ ∠OBC = 90° − 40° = 50° By degree measure theorem
∠AOC = 2∠OBC
⇒ x = 2 × 50° = 100°
(ix) In ΔDAB, by angle sum property
∠ADB + ∠DAB + ∠ABD = 180°
⇒ 32° + ∠DAB + 50° = 180°
⇒ ∠DAB = 180° − 32° − 50°
⇒ ∠DAB = 98°
Now, ∠OAB + ∠DCB = 180° [Opposite angles of cyclic quadrilateral]
⇒ 98° + x = 180°
⇒ x = 180° − 98° = 82°
(x) We have,
∠BAC = 35°
∠BDC = ∠BAC = 35° [Angle in same segment]
In ΔBCD, by angle sum property
⇒ 35° + x + 65° = 180°
⇒ x = 180° − 35° − 65° = 80°
(xi) We have
∠ABD = 40°
∠ACD = ∠ABD = 40° [Angle in same segment]
In ΔPCD, by angle sum property
∠PCD + ∠CPO + ∠PDC = 180°
⇒ 40° + 110° + x = 180°
⇒ x = 180° − 150°
⇒ x = 30°
(xii) Given that,
∠BAC = 52°
Then ∠BDC = ∠BAC = 52° [Angle in same segment]
Since OD = OC
Then ∠ODC = ∠OCD [Opposite angle to equal radii]
⇒ x = 52°
O is the circumference of the triangle ABC and OD is perpendicular on BC. Prove that ∠BOD = ∠A.
Given O is the circum centre of triangle ABC and OD⊥BC
To prove ∠BOD = 2∠A
Proof:
In ΔOBD and ΔOCD
∠ODB = ∠ODC [Each 90°]
OB = OC [Radius of circle]
OD = OD [Common]
Then ΔOBD ≅ ΔOCD [By RHS Condition].
∴ ∠BOD = ∠COD .... (i) [PCT].
⇒ 2∠BOD = 2∠BAC [By using (i)]
⇒ ∠BOD = ∠BAC.
In figure, O is the centre of the circle, BO is the bisector of ∠ABC. Show that AB = AC.
Given, BO is the bisector of ∠ABC
To prove AB = BC
Since, BO is the bisector of ∠ABC.
Then, ∠ABO = ∠CBO ... (i)
Since, OB = OA [Radius of circle]
Then, ∠ABO = ∠DAB... (ii) [opposite angles to equal sides]
Since OB = OC [Radius of circle]
Then, ∠OAB = ∠OCB... (iii) [opposite angles to equal sides]
Compare equations (i), (ii) and (iii)
∠OAB = ∠OCB ... (iv)
In ΔOAB and ΔOCB
∠OAB = ∠OCB From (iv)]
∠OBA = ∠OBC [Given]
OB = OB [Common]
Then, ΔOAB ≅ ΔOCB [By AAS condition]
∴ AB = BC [CPCT]
In figure, O is the centre of the circle, then prove that ∠x = ∠y + ∠z.
We have,
∠3 = ∠4 [Angles in same segment]
∴ ∠x = 2∠3 [By degree measure theorem]
⇒ ∠x = ∠3 + ∠3
⇒ ∠x = ∠3 + ∠4 ... (i) [∠3 = angle 4]
But ∠y = ∠3 + ∠1 [By exterior angle property]
⇒ ∠3 = ∠y − ∠1 .... (ii)
from (i) and (ii)
∠x = ∠y − ∠1 + ∠4
⇒ ∠x = ∠y + ∠4 − ∠1
⇒ ∠x = ∠y + ∠z + ∠1 − ∠1 [By exterior angle property]
⇒ ∠x = ∠y + ∠z
In figure, O and O' are centers of two circles intersecting at B and C. ACD is a straight line, find x.
⇒ 130° = 2∠ACB
⇒ ∠ACB = 130°/2 = 65°
∴ ∠ACB + ∠BCD = 180° [Liner a pair of angles]
⇒ 65° + ∠BCD = 180°
⇒ ∠BCD = 180° − 65° = 115° By degree measure theorem reflex
∠BOD = 2∠BCD
⇒ reflex ∠BOD = 2 × 115° = 230°
Now, reflex ∠BOD + ∠BO′D = 360° [Complex angle]
⇒ 230° + x = 360°
⇒ x = 360° − 230°
∴ x = 130°
In figure, O is the centre of a circle and PQ is a diameter. If
∠ROS = 40°, find ∠RTS.
Since PQ is diameter
Then,
∠PRQ = 90° [Angle in semicircle]
∴ ∠PRQ + ∠TRQ = 180° [Linear pair of angle]
900 + ∠TRQ = 180°
∠TRQ = 180° − 90° = 90°.
∠ROS = 2∠RQS
⇒ 40° = 2∠RQS
⇒ ∠RQS = 40°/2 = 20°
In ΔRQT, by Angle sum property
∠RQT + ∠QRT + ∠RTS = 180°
⇒ 20° + 90° + ∠RTS = 180°
⇒ ∠RTS = 180° − 20° − 90° = 70°
In figure, if ∠ACB = 40°, ∠DPB = 120°, find ∠CBD.
∠ACB = 40°; ∠DPB = 120°
∴ ∠APB = ∠DCB = 40° [Angle in same segment]
In ΔPOB, by angle sum property
∠PDB + ∠PBD + ∠BPD = 180°
⇒ 40° + ∠PBD + 120° = 180°
⇒ ∠PBD = 180° − 40° − 120°
⇒ ∠PBD = 20°
∴ ∠CBD = 20°
A chord of a circle is equal to the radius of the circle. Find the angle subtended by the chord at a point on the minor arc and also at a point on the major arc.
Radius OA = Chord AB
⟹ OA = OB = AB
Then triangle OAB is an equilateral triangle.
∴ ∠AOB = 60° [one angle of equilateral triangle]
⇒ 60° = 2∠APB
⇒ ∠APB = 60°/2 = 30°
Now, ∠APB + ∠AQB = 180° [opposite angles of cyclic quadrilateral]
⇒ 300 + ∠AQB = 180°
⇒ ∠AQB = 180° − 30° = 150°.
Therefore, Angle by chord AB at minor arc = 150°
Angle by chord AB at major arc = 30°
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