A chord of a circle is equal to the radius of the circle. Then the angle subtended by the chord at a point on the minor arc and the also at a point on the major arc are
respectively

120^{o} & 60^{o}

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130^{o} & 50^{o}

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140^{o} & 40^{o}

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150^{o} & 30^{o}

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Solution

The correct option is B $150^{o} & 30^{o}$
$G i v e n - O i s t h e c e n t r e o f a c i r c l e w i t h A B a s a c h o r d . A B = t h e r a d i u s O A o r O B . A B s u b t e n d s ∠ A C B t o t h e m a j o r a r c A P R B a t C a n d ∠ A D B t o t h e m i n o r a r c A D B a t D T o f i n d o u t - ∠ A D B = ? S o l u t i o n - O A = O B = A B (g i v e n) . ∴ Δ O A B i s e q i l a t e r a l . i . e ∠ A O B = 60^{o} a n d r e f l e x ∠ A O B = 360^{o} - 60^{o} = 300^{o} . S o ∠ A D B = \frac{1}{2} r e f l e x ∠ A O B = \frac{1}{2} \times 300^{o} = 150^{o} . (s i n c e t h e a n g l e, s u b t e n d e d b y a c h o r d o f a c i r c l e a t t h e c e n t r e, i s d o u b l e t h e a n g l e s u b t e n d e d b y t h e s a m e c h o r d a t t h e c i r c u m f e r e n c e o f t h e c i r c l e .) . . . . . . (i) B y t h e s a m e a r g u m e n t a s (i), w e g e t, ∠ A C B = \frac{1}{2} ∠ A O B = \frac{1}{2} \times 60^{o} = 30^{o} . s o a n g l e s, s u b t e n d e d b y A B t o t h e m i n o r a r c & t h e m a j o r a r c, a r e r e s p e c t i v e l y 150^{o} & 30^{o} . A n s - O p t i o n D .$

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A chord of a circle is equal to the radius of the circle. Then the angle subtended by the chord at a point on the minor arc and the also at a point on the major arc arerespectively

A chord of a circle is equal to the radius of the circle. Then the angle subtended by the chord at a point on the minor arc and the also at a point on the major arc are
respectively