Selina Concise Class 9 Maths Chapter 17 Circles Exercise 17C Solutions
EXERCISE – 17C
(Q1) In the given figure, an equilateral triangle ABC is inscribed in a circle with centre O. Find
(i) ∠BOC
(ii) ∠OBC
Solution:
Angle subtend at centre by equilateral triangle.
= 360/3
∠BOC => 120°
In △BOC,
∠BOC + ∠OBC + COCB = 180°
120° + x + x = 180°
2x = 180° – 120°
2x = 60°
x = 60/2
x = 30°
∴ ∠OBC = 30°
(Q2) In the given figure, a square is inscribed in a circle with centre O. find:
(i) ∠BOC
(ii) ∠OCB
(iii) ∠COD
(iv) ∠BOD
Is BD a diameter of the circle?
Solution:
Each angle subtend by square at centre = 360°/4
∠AOB = ∠AOD = ∠COD = ∠BOC = 90°
In △BOC,
∠BOC + x + x = 180°
90° + 2x = 180°
2x = 180° – 90°
2x = 90°
x = 90°/2
x = 45°
∴ ∠OBC = 45°
(iv) ∠BOC + ∠COD
90° + 90°
180°
Hence, BD is a diameter.
(Q3) In the given figure, AB is a side of regular pentagon and BC is a side of regular hexagon.
(i) ∠AOB
(ii) ∠BOC
(iii) ∠AOC
(iv) ∠OBA
(v) ∠OBC
(vi) ∠ABC
Solution:
Angle subtend by AB at centre = 360°/5
∠AOB = 72°
(ii) Angle subtend by BC at centre = 360°/6
= 60°
= ∠BOC)
(iii) ∠AOC = ∠AOB + ∠BOC
= 72° + 60°
= 132°
(iv) In △AOB, ∠AOB + x + x = 180
72° + 2x = 180°
2x = 180 – 72°
2x = 108
x = 108/2
x = 54°
∠OBA = 54°
(v) In △BOC, ∠BOC + y + y = 180°
60° + 2y = 180°
2y = 180° – 60°
2y = 120°
y = 120°/2
y = 60°
∴ ∠OBC = 60°
(vi) ∠ABC = ∠OBA + ∠OBC
= 54° + 60°
∠ABC = 114°
(Q4) In the given figure, arc AB and arc BC are equal in length. If ∠AOB = 48°, find:
(i) ∠BOC
(ii) ∠OBC
(iii) ∠OBC
(iv) ∠OAC
Solution:
(i) ∠BOC = 48°
(ii) In △BOC,
∠BOC + ∠OBC + ∠OCB = 180° (The sum of all interior angles = 180°)
48° + x + x = 180°
2x = 180° – 48°
2x = 132
x = 132/2
x = 66°
(iii) ∠AOC = ∠AOB + ∠BOC
= 48° + 48°
= 96°
(iv) In △AOC,
∠AOC + ∠OAC + ∠OCA = 180°
96 + y + y = 180°
2y = 180° – 96°
2y = 84
y = 84/2
y = 42°
∠OAC = 42°
(Q5) In the given figure, the lengths of arcs AB and BC are in the ratio 3:2. If ∠AOB = 96°, find
(i) ∠BOC
(ii) ∠ABC
Solution:
Given:
AB/BC = 3/2 = 3x/2x
∠AOB = 96°
Find: (i) ∠BOC
(ii) ∠ABC
Solution:
∠AOB = 96°
3x = 96
x = 96/3
x = 32°
(i) 2x = 2 × (32)
∠BOC = 64°
(ii) In △AOB,
∠AOB + x + x = 180°
96° + 2x = 180°
2x = 180° – 96°
2x = 84
x = 84/2
x = 42°
In △AOC,
∠BOC + y + y = 180°
64° + 2y = 180°
2y = 180 – 64
2y = 116
y = 116/2
y = 58°
∠ABC = ∠OBA + ∠OBC
= 42° + 58°
∠ABC = 100°
Here is your solution of Selina Concise Class 9 Maths Chapter 17 Circles Exercise 17C
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