ANGLE SUBTENDED BY AN ARC AT THE CENTER OF THE CIRCLE

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.

O is the centre of the circle. AXB is the arc. AOB is the angle  subtended by the arc AXB at the centre. <ACB is the angle subtended by the arc AXB at a point on the remaining part of the circle

To prove : AOB  =  2ACB

Construction : Join CO and produce it to D

(i) OA  =  OC     (Radii)

(ii) OCA  =  OAC 

(angles opposite to equal sides are equal.)

(iii) In ΔAOC 

<AOD  =  OCA + OAC

(Exterior angles of a triangle  =  Sum of the interior opposite angles)

(iv) AOD  =  OCA + OCA

(substituting <OAC by <OCA)

(v) AOD  =  2OCA (by addition)

(vi) Similarly in triangle BOC

BOD  =  2OCB

(vii) AOD + BOD  =  2OCA + 2OCB

=  2(OCA + OCB) 

(AOD + BOD  =  AOB and OCA + OCB  =  ACB)

(viii) AOB  =  2ACB

Example 1 :

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 2 :

Find the value of x in the following figure.

Solution :

reflex AOB  =  2ACB

x = 2 ⋅ 100°  =  200°

Example 3 :

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 4 :

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°) 

=  90°

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