Important for :

1

D

OA = OB = OC = Circum-radius

In ΔABC, we know that

∠ABC + ∠BCA + ∠BAC = 180°

∠BAC = 180° – 70° – 40° = 70°

**Note :** The angle subtended by an arc of a circle at the centre is double the angle subtended by it at any point on the remaining part of the circle.

∴ ∠BOC = 2 × ∠BAC = 2 × 70° = 140°

In ΔABC, we know that

∠ABC + ∠BCA + ∠BAC = 180°

∠BAC = 180° – 70° – 40° = 70°

∴ ∠BOC = 2 × ∠BAC = 2 × 70° = 140°

Hence, option D is correct.

2

B

∴ Exterior ∠BOC = 2 × ∠BAC = 2 × 90° = 180°

∴ ∠BOC = 360° – Exterior ∠BOC = 360° – 180° = 180°

OA = OB = OC = r cm (radii)

AB = AC

∴ ∠AOB = ∠AOC = 90°

In ΔAOB, By pythagoras theorem

AB

(5√2)

r

r = 5 cm

Hence, option B is correct.

3

D

∠ONY = 50° and ∠OMY = 15°

In ΔONY,

ON = OY (radii)

∠OYN = ∠ONY = 50°

∴ ∠NOY = 180° – ∠ONY – ∠OYN = 180° – 50° – 50° = 80°

In ΔOMY,

OM = OY (radii)

∠OYM = ∠OMY = 15°

∴ ∠MOY = 180° – ∠OMY – ∠OYM = 180° – 15° – 15° = 150°

∴ ∠MON = ∠MOY – ∠NOY = 150° – 80° = 70°

Hence, option D is correct.

4

A

AB = a metre and CD = b metre

∠AOB = 60° and ∠COD = 90°

In ΔCOD, By pythagoras theorem

CD

b

In ΔAOB,

OA = OB

∴ ∠ABO = ∠OAB

∠AOB + ∠ABO + ∠OAB = 180°

60° + ∠OAB + ∠OAB = 180°

2∠OAB = 180° – 60° = 120°

∠OAB = 60° = ∠ABO

∴ ΔAOB is an equilateral triangle.

OA = OB = AB ⇒ a = r

From equation (i),

b = √2r

b = √2a

Hence, option A is correct.

5

C

∠POQ = ∠POR = 90°

OQ = OP = OR [

∴ ∠OQP = ∠OPQ = ∠ORP = ∠OPR

In ΔPOQ, we know that

∠POQ + ∠OQP + ∠OPQ = 180°

90° + ∠OPQ + ∠OPQ = 180°

2∠OPQ = 180° – 90° = 90°

∠OPQ = 45°

Similarly in ΔPOR, we get

∠ORP = 45°

∴ ∠QPR = ∠OPQ + ∠ORP = 45° + 45° = 90°

Hence, option C is correct.