ANGLE SUBTENDED AT THE CENTER OF A CIRCLE

The angle at the centre of a circle is twice the angle on the circle subtended by the same arc

Find, giving reasons, the values of the pronumerals in the following:

Example 1 :

Solution :

<AOB  =  128 (Given)

<AOB  =  2<ABC

Angle subtended at the center is twice the angle subtended by the same arc.

2x  =  128

x  =  128/2

x  =  64

Example 2 :

Solution :

<AOC  =  2<ABC

2<ABC  =  80

<ABC  =  40

In triangle ABC,

AB  =  BC

So, <BAC  =  <BCA  =  x

<ABC + <BCA + <BAC  =  180

40+x+x  =  180

2x  =  180-40

2x  =  140

x  =  70

Example 3 :

Solution :

<COB  =  x

OA  =  OB  =  radius

<OAC  =  <OCA  =  33

In triangle AOC

<OAC + <OCA + <AOC  =  180

33+33+<AOC  =  180

<AOC  =  180-66

<AOC  =  114

<AOC + <COB  =  180

114+x  =  180

x  =  180-114

x  =  66

Example 4 :

Solution :

<AOD  =  88

2<ABC  =  88

<ABC  =  44

In triangle ABC,

<CBA + <BAC + <ACB  =  180

44+50+<ACB  =  180

94+<ABC  =  180

<ABC  =  180-94

<ABC  =  86

<BCA + <ACD  =  180

86+x  =  180

x  =  180-86

x  =  94

Example 5 :

Solution :

<AOC  =  100

2<ABC  =  <AOB

2<ABC  =  100

<ABC  =  50

ABCO is a quadrilateral.

<OAB + <ABC + <BCO + <AOC  =  360

x + 50 + x + 260  =  360

2x + 310  =  360

2x  =  360-310

2x  =  50

x  =  25

Example 6 :

Solution :

<OAB  =  45, <OBA  =  45, <AOB  =  90

<ADB  =  90

<AOB  =  90 (2 times of <ADB because angle at the center)

<ADB  =  90/2

<ADB  =  45

After having gone through the stuff given above, we hope that the students would have understood, how to order fractions and decimals. 

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