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In the given figure, 0 is the centre of ...

In the given figure, 0 is the centre of a circle and arc ABC subtends an angle of `130^@` at 0. AB is extended to P. Then, `angle PBC` is equal to

A

`25^@`

B

`40^@`

C

`65^@`

D

`75^@`

Text Solution

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The correct Answer is:
C
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In the given figure, O is the centre of a circle and arc ABC subtends an angle of 130^(@) at O. AB is extended to P, Then, anglePBC is equal to

In the given figure, O is the centre of a circle and angle ADC=130^@ .If angle BAC=x^@ , Find the value of x.

In a circle with centre O, an arc ABC subtends an angle of 136^@ at the centre of the circle. Chord AB is produced to point P. Then angle CBP is equal to : केंद्र Oवाले एक वृत्त में, चाप ABC वृत्त के केंद्र पर 136^@ का कोण अंतरित करता है | जीवा AB को बिंदु P तक बढ़ाया जाता है | तो angle CBP का मान क्या होगा ?

In a circle with centre O, an arc ABC subtends an angle of 132^@ at the centre of the circle. Chord AB is produced to point P. Then angle CBP is equal to : केंद्र Oवाले एक वृत्त में, चाप ABC वृत्त के केंद्र पर 132^@ का कोण अंतरित करता है | जीवा AB को बिंदु P तक बढ़ाया जाता है | तो angle CBP का मान क्या होगा ?

In a circle with centre O, an arc ABC subtends an angle of 140^@ at the centre of the circle. Chord AB is produced to point P. Then angle CBP is equal to : केंद्र Oवाले एक वृत्त में, चाप ABC वृत्त के केंद्र पर 140^@ का कोण अंतरित करता है | जीवा AB को बिंदु P तक बढ़ाया जाता है | तो angle CBP का मान क्या होगा ?

In a circle with centre O, an arc ABC subtends an angle of 110^@ at the centre of the circle. Chord AB is produced to point P. Then angle CBP is equal to : केंद्र Oवाले एक वृत्त में, चाप ABC वृत्त के केंद्र पर 110^@ का कोण अंतरित करता है | जीवा AB को बिंदु P तक बढ़ाया जाता है | तो angle CBP का मान ज्ञात करें |

In figure, if 0 is the centre of a circle, PQ is a chord and the tangent PR at P makes an angle of 50^(@) with PQ, then anglePOQ is equal to

In a circle with centre O. an arc ABC subtends an angle of 136^(@) at the centre of the circle. The chord AB is produced to a point P. Then angle CBP is equal to :

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