ANGLE SUBTENDED BY AN ARC AT THE CENTRE IS DOUBLE PROOF

About "Angle subtended by an arc at the centre is double proof"

Angle subtended by an arc at the centre is double proof :

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 = 2 <ACB

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  =  2 <OCA (by addition)

(vi) Similarly in triangle BOC

<BOD  =  2 <OCB

(vii) <AOD + <BOD  =  2 <OCA + 2<OCB

=  2(<OCA + <OCB) 

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

(viii) <AOB = 2 <ACB

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  =  2<ACB

<ACB  =  (1/2) <AOB

  =  (1/2)  80°

 =  40°

Example 2 :

Find the value of x in the following figure.

Solution :

reflex <AOB  =  2 <ACB

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  =  2 <ACB

x  =  2 (45°) 

=  90°

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