(1) Find the value of x in the following figure. | |

(2) Find the value of x in the following figure. | |

(3) Find the value of x in the following figure. | |

(4) Find the value of x in the following figure. | |

(5) Find the value of x in the following figure. | |

(6) Find the value of x in the following figure. | |

(7) Find the value of x in the following figure. | |

(8) Find the value of x in the following figure. | |

(9) Find the value of x in the following figure. | |

(10) Find the value of x in the following figure. |

**Question 1 :**

Find the value of x in the following figure.

**Solution :**

According to the theorem,

∠BOC = 2∠BAC -----(1)

∠BOC = 2x

∠BOC + ∠BOA + ∠AOC = 360°

∠BOC + 120 + 90 = 360

∠BOC + 210 = 360

∠BOC = 360 - 210

∠BOC = 150°

From this we may find the value of ∠BAC. That is x

From (1)

∠BOC = 2∠BAC

150 = 2∠BAC

∠BAC = 150/2

= 75°

So, the value of x is 75.

**Question 2 :**

Find the value of x in the following figure.

**Solution :**

The angle in a semicircle is right angle.

∠ACB = 90°

Sum of interior angels in a triangle is 180°.

∠CAB + ∠ACB + ∠ABC = 180°

x + 90 + 35 = 180

x + 125 = 180

x = 180 - 125

x = 55

**Question 3 :**

Find the value of x in the following figure.

OA = OB = OC ( radius )

∠OCA = ∠OAC = 30°

∠OBA = ∠OAB = 25°

∠BOC = 2∠BAC

∠BAC = ∠BAO + ∠OAC

= 25 + 30

= 55°

∠BOC = 2(55) = 110°

So, the value of x is 110.

**Question 4 :**

Find the value of x in the following figure.

**Solution :**

reflex ∠AOC = 360 - 130 = 230°

2∠ABC = reflex of ∠AOC

Reflex of ∠AOC = 2x

x = Reflex of ∠AOC / 2

= 230/2

= 115°

So, the value of x is 115.

**Question 5 :**

Find the value of x in the following figure.

**Solution :**

In triangle DBC,

∠DBC + ∠DCB + ∠CDB = 180°

∠CDB = 90°

50 + x + 90 = 180

140 + x = 180°

x = 180 - 140

x = 40

So, the value of x is 40.

**Question 6 :**

Find the value of x in the following figure.

∠AOB = 2∠ACB

∠AOB = 2(48) = 96

In triangle AOB,

∠OAB + ∠ABO + ∠BOA = 180

OA = OB = OC = Radius

∠OAB = ∠OBA = x

x + x + 96 = 180

2x + 96 = 180

2x = 180 - 96

2x = 84

x = 84/2

x = 42

So, the value of x is 42.

**Question 7 :**

Find the value of x in the following figure.

**Solution :**

Using the theorem 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.

∠ACB = (1/2)∠AOB

x° = (1/2) ⋅ 80°

x° = 40°

So, the value of x is 40.

**Question 8 :**

Find the value of x in the following figure.

**Solution :**

reflex ∠OB = 2∠ACB

x = 2 ⋅ 100° = 200°

**Question 9 :**

Find the value of x in the following figure.

∠ABC + ∠BCA + ∠CAB = 180°

56° + 90° + ∠CAB = 180°

(∠BCA = angle on a semicircle = 90°)

∠CAB = 180° - 146°

x° = 34°

So, the value of x is 34.

**Question 10 :**

Find the value of x in the following figure.

OA = OB = OC ( radius )

∠OCA = ∠OAC = 25°

∠OBC = ∠OCB = 20°

∠ACB = ∠OCA + ∠OCB

= 25° + 20° = 45°

∠AOB = 2∠ACB

x° = 2(45°)

x° = 90°

So, the value of x is 90.

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