Important for :

1

C

Let radius be 'r' and ∠POS = x°

ΔOQR isoceles ∴∠QOR = 30°

∴ ∠OQR = 120° (Sum of all angles of ΔOQR = 180°)

∴ ∠OQP = 60° (Supplementary agnle)

ΔOPQ isoceles since OP = OQ = r

∴ ∠OQP = 60° = ∠OQP

∴ ∠ POQ = 60° = [Sum of all angle of Δ = 180° ]

Now SOR is a straight line

∴ x + 60° + 30° = 180°

∴ x = 90°

Hence, option (C) is correct.

2

B

∠ PQS = 60°

∠QO'R = 130°

∠ QPR = | 1 | × 130° = 65° |

2 |

⇒ ∠ QRP = 180° – 60° – 65° = 55°

⇒ ∠PO'Q = 110°

In Δ QO'R

QO' = O'R

⇒ ∠O'QR = ∠O'RQ = 25°

∵ ∠O'QR + ∠O'RQ = 50°

⇒ ∠PQO' + ∠QPO' = 35°

∵ ∠PQO' + ∠QPO' = 70°

Similarly, ∠O'PR = 30°

∴ ∠RPS = 35°

Hence, option (B) is correct.

3

B

∠AOB = 48°So, ∠ACB = | 1 | ∠AOB |

2 |

= | 1 | × 48° = 24° |

2 |

(As angles made by same arc AB)

Given AC and OB intersect each other at right angle.

∠ CQB = 90°

∠ CBQ = 180° – (90° + 24°) = 66°

so , ∠ OBC = 66°

Hence, option B is correct.

4

C

∠APB = 90°

AB = Diameter = Hypotenous of triangle APB

As, the angle of semicircle is right angle

so, the circumcentre lies on midpoint of hypoteneous

Hence, option (C) is correct.

5

B

In ∆ AOD :

OA = OD (radius)

∠ AOD = 90 ( as OD is perpendicular to AB )

So ∆ AOD is isosceles having OA and OD sides equal and one angle as 90

So the remaining wo angles are 45 each

Hence ∠ BAD = 45°

Therefore, option (B) is correct.

Therefore, option (B) is correct.