RD Chapter 16 Circles Ex VSAQS Solutions
Question - 1 : - In the figure, two circles intersect at A and B. The centre of the smaller circle is O and it lies on the circumference of the larger circle. If ∠APB = 70°, find ∠ACB.
Answer - 1 : -
Arc AB subtends ∠AOB at the centre and ∠APB at the remaining part of the circle
∴ ∠AOB = 2∠APB = 2 x 70° = 140°
Now in cyclic quadrilateral AOBC,
∠AOB + ∠ACB = 180° (Sum of the angles)
⇒ 140° +∠ACB = 180°
⇒ ∠ACB = 180° – 140° = 40°
∴ ∠ACB = 40°
Question - 2 : - In the figure, two congruent circles with centre O and O’ intersect at A and B. If ∠AO’B = 50°, then find ∠APB.
Answer - 2 : -
Two congruent circles with centres O and O’ intersect at A and B
∠AO’B = 50°
∵ OA = OB = O’A = 04B (Radii of the congruent circles)
Question - 3 : - In the figure, ABCD is a cyclic quadrilateral in which ∠BAD = 75°, ∠ABD = 58° and ∠ADC = IT, AC and BD intersect at P. Then, find ∠DPC.
Answer - 3 : -
∵ ABCD is a cyclic quadrilateral,
∴ ∠BAD + ∠BCD = 180°
⇒ 75° + ∠BCD – 180°
⇒ ∠BCD = 180°-75°= 105° and ∠ADC + ∠ABC = 180°
⇒ 77° + ∠ABC = 180°
⇒ ∠ABC = 180°-77°= 103°
∴ ∠DBC = ∠ABC – ∠ABD = 103° – 58° = 45°
∵ Arc AD subtends ∠ABD and ∠ACD in the same segment of the circle 3
∴ ∠ABD = ∠ACD = 58°
∴ ∠ACB = ∠BCD – ∠ACD = 105° – 58° = 47°
Now in ∆PBC,
Ext. ∠DPC = ∠PBC + ∠PCB
=∠DBC + ∠ACB = 45° + 47° = 92°
Hence ∠DPC = 92°
Question - 4 : - In the figure, if ∠AOB = 80° and ∠ABC = 30°, then find ∠CAO.
Answer - 4 : -
In the figure, ∠AOB = 80°, ∠ABC = 30°
∵ Arc AB subtends ∠AOB at the centre and
∠ACB at the remaining part of the circle
∴ ∠ACB = 
∠AOB =x 80° = 40° In ∆OAB, OA = OB
∴ ∠OAB = ∠OBA
But ∠OAB + ∠OBA + ∠AOB = 180°
∴ ∠OAB + ∠OBA + 80° = 180°
⇒ ∠OAB + ∠OAB = 180° – 80° = 100°
∴ 2∠OAB = 100°
⇒ ∠OAB = 100/2 = 50°
Similarly, in ∆ABC,
∠BAC + ∠ACB + ∠ABC = 180°
∠BAC + 40° + 30° = 180°
⇒ ∠BAC = 180°-30°-40°
= 180°-70°= 110°
∴ ∠CAO = ∠BAC – ∠OAB
= 110°-50° = 60°
Question - 5 : - In the figure, A is the centre of the circle. ABCD is a parallelogram and CDE is a straight line. Find ∠BCD : ∠ABE.
Answer - 5 : -
In the figure, ABCD is a parallelogram and
CDE is a straight line
∵ ABCD is a ||gm
∴ ∠A = ∠C
and ∠C = ∠ADE (Corresponding angles)
⇒ ∠BCD = ∠ADE
Similarly, ∠ABE = ∠BED (Alternate angles)
∵ arc BD subtends ∠BAD at the centre and
∠BED at the remaining part of the circle
Question - 6 : - In the figure, AB is a diameter of the circle such that ∠A = 35° and ∠Q = 25°, find ∠PBR.
Answer - 6 : -
In the figure, AB is the diameter of the circle such that ∠A = 35° and ∠Q = 25°, join OP.
Arc PB subtends ∠POB at the centre and
∠PAB at the remaining part of the circle
∴ ∠POB = 2∠PAB = 2 x 35° = 70°
Now in ∆OP,
OP = OB radii of the circle
∴ ∠OPB = ∠OBP = 70° (∵ ∠OPB + ∠OBP = 140°)
Now ∠APB = 90° (Angle in a semicircle)
∴ ∠BPQ = 90°
and in ∆PQB,
Ext. ∠PBR = ∠BPQ + ∠PQB
= 90° + 25°= 115°
∴ ∠PBR = 115°
Question - 7 : - In the figure, P and Q are centres of two circles intersecting at B and C. ACD is a straight line. Then, ∠BQD =
Answer - 7 : -
In the figure, P and Q are the centres of two circles which intersect each other at C and B
ACD is a straight line ∠APB = 150°
Arc AB subtends ∠APB at the centre and
∠ACB at the remaining part of the circle
∴ ∠ACB = ∠APB = x 150° = 75° But ∠ACB + ∠BCD = 180° (Linear pair)
⇒ 75° + ∠BCD = 180°
∠BCD = 180°-75°= 105°
Now arc BD subtends reflex ∠BQD at the centre and ∠BCD at the remaining part of the circle
Reflex ∠BQD = 2∠BCD = 2 x 105° = 210°
But ∠BQD + reflex ∠BQD = 360°
∴ ∠BQD+ 210° = 360°
∴ ∠BQD = 360° – 210° = 150°
Question - 8 : - In the figure, if O is circumcentre of ∆ABC then find the value of ∠OBC + ∠BAC.
Answer - 8 : -
In the figure, join OC
∵ O is the circumcentre of ∆ABC
∴ OA = OB = OC
∵ ∠CAO = 60° (Proved)
∴ ∆OAC is an equilateral triangle
∴ ∠AOC = 60°
Now, ∠BOC = ∠BOA + ∠AOC
= 80° + 60° = 140°
and in ∆OBC, OB = OC
∠OCB = ∠OBC
But ∠OCB + ∠OBC = 180° – ∠BOC
= 180°- 140° = 40°
⇒ ∠OBC + ∠OBC = 40°
∴ ∠OBC =
= 20° ∠BAC = OAB + ∠OAC = 50° + 60° = 110°
∴ ∠OBC + ∠BAC = 20° + 110° = 130°
Question - 9 : - In the AOC is a diameter of the circle and arc AXB = 1/2 arc BYC. Find ∠BOC.
Answer - 9 : -
In the figure, AOC is diameter arc AxB = arc BYC 1 ∠AOB = ∠BOC ⇒ ∠BOC = 2∠AOB
But ∠AOB + ∠BOC = 180°
⇒ ∠AOB + 2∠AOB = 180°
⇒ 3 ∠AOB = 180°
∴ ∠AOB =
= 60° ∴ ∠BOC = 2 x 60° = 120°
Question - 10 : - In the figure, ABCD is a quadrilateral inscribed in a circle with centre O. CD produced to E such that ∠AED = 95° and ∠OBA = 30°. Find ∠OAC.
Answer - 10 : -
In the figure, ABCD is a cyclic quadrilateral
CD is produced to E such that ∠ADE = 95°
O is the centre of the circle
∵ ∠ADC + ∠ADE = 180°
⇒ ∠ADC + 95° = 180°
⇒ ∠ADC = 180°-95° = 85°
Now arc ABC subtends ∠AOC at the centre and ∠ADC at the remaining part of the circle
∵ ∠AOC = 2∠ADC = 2 x 85° = 170°
Now in ∆OAC,
∠OAC + ∠OCA + ∠AOC = 180° (Sum of angles of a triangle)
⇒ ∠OAC = ∠OCA (∵ OA = OC radii of circle)
∴ ∠OAC + ∠OAC + 170° = 180°
2∠OAC = 180°- 170°= 10°
∴ ∠OAC = 10/2 = 5°