ANGLE SUBTENDED BY AN ARC AT THE CENTRE PRACTICE

About "Angle subtended by an arc at the centre practice"

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.

Angle subtended by an arc at the centre practice - Examples

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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