In the given figure POQ is a diameter of a circle with centre O and PQRS is a cyclic quadrilateral. SQ is joined. If $∠ R = 138^{o}$ . Then $∠ P Q S$ is

90^{o}

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42^{o}

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48^{o}

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38^{o}

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Solution

The correct option is C $48^{o}$
$G i v e n - P O Q i s t h e d i a m e t e r o f a c i r c l e w i t h c e n t r e O . P Q R S i s a q u a d r i l a t e r a l i n s c r i b e d i n t h e g i v e n c i r c l e . S Q h a s b e e n j o i n e d . ∠ Q R S = 138^{o} . T o f i n d o u t - ∠ P Q S = ? S o l u t i o n - P Q R S i s a c y c l i c q u a d r i l a t e r a l . ∴ ∠ Q P S + ∠ Q R S = 180^{o} (s i n c e t h e s u m o f t h e o p p o s i t e a n g l e s o f a c y c l i c q u a d r i l a t e r a l i s 180^{o}) . ⟹ ∠ Q P S = 180^{o} - ∠ Q R S = 180^{o} - 138^{o} = 42^{o} . N o w P Q i s t h e d i a m e t e r w h i c h s u b t e n d s ∠ Q P S t o t h e c i r c u m f e r e n c e o f t h e g i v e n c i r c l e a t S . ∴ ∠ Q P S = 90^{o} s i n c e i t i s t h e a n g l e i n a s e m i c i r c l e . S o i n Δ P Q S w e h a v e ∠ Q S P + ∠ Q P S + ∠ P Q S = 180^{o} (a n g l e s u m p r o p e r t y o f t r i a n g l e s) . ⟹ ∠ P Q S = 180^{o} - (∠ Q S P + ∠ Q P S) = 180^{o} - (90^{o} + 42^{o}) = 48^{o} . A n s - O p t i o n C .$

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