**Angle subtended by an arc at the centre practice :**

Here we are going to practice questions on angle subtended by an arc.

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.

**Example 1 :**

Find the value of x in the following figure.

**Solution :**

According to the theorem,

<BOC = 2 <BAC --(1)

<BOC = 2 x

<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°

Hence the value of x is 75°.

**Example 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°

**Example 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°

Hence the value of x is 110°

**Example 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°

Hence the value of x is 115°

**Example 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°

Hence the value of x is 40°.

**Example 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°

Hence the value of x is 42°.

**Example 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.

(i) <AOB = 2<ACB

<ACB = (1/2) <AOB

= (1/2) ⋅ 80°

= 40°

**Example 8 :**

Find the value of x in the following figure.

**Solution :**

reflex <AOB = 2 <ACB

x = 2 ⋅ 100° = 200°

**Example 9 :**

Find the value of x in the following figure.

<ABC + <BCA + <CAB = 180°

56° + 90° + <CAB = 180°

(a <BCA = angle on a semicircle = 90°)

<CAB = 180° - 146°

x = 34°

**Example 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°)

= 90°

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