# ANGLE SUBTENDED BY AN ARC AT THE CENTRE PRACTICE

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.

## Solved Examples

Example 1 :

Find the value of x in the following figure.

Solution :

According to the theorem,

BOC  =  2BAC -----(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  =  2BAC

150  =  2BAC

BAC  =  150/2

=  75°

So, 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  =  2BAC

BAC  =  BAO + OAC

=  25 + 30

=  55°

BOC  =  2(55)  =  110°

So, the value of x is 110°.

Example 4 :

Find the value of x in the following figure.

Solution :

reflex AOC  =  360 - 130  =  230°

2ABC  =  reflex of AOC

Reflex of AOC  =  2x

x  =  Reflex of AOC / 2

=  230/2

=  115°

So, 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°

So, the value of x is 40°.

Example 6 :

Find the value of x in the following figure.

AOB  =  2ACB

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°.

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  =  2ACB

ACB  =  (1/2)AOB

=  (1/2)  80°

=  40°

Example 8 :

Find the value of x in the following figure.

Solution :

reflex AOB  =  2ACB

x  =  2 ⋅ 100°  =  200°

Example 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°

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  =  2ACB

x  =  2(45°)

x  =  90°

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