In the given figure $P Q$ is a diameter of a circle with centre $O$ and $P T$ is a tangent at $P$ . $Q T$ meets the circle at $R$ . If $∠ P O R = 72^{o}$ , $∠ P T R$ is

118^{o}

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36^{o}

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54^{o}

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26^{o}

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Solution

The correct option is B $54^{o}$
$G i v e n - P Q i s a d i a m e t e r o f a c i r c l e w i t h c e n t r e O . P T i s a t a n g e n t t o t h e c i r c l e f r o m T a t P . Q T i n t e r s e c t s t h e c i r c l e a t R . ∠ P O R = 72^{o} . T o f i n d o u t - ∠ P T R = ? S o l u t i o n - T h e d i a m e t e r P Q o r t h e r a d i u s O P m e e t s t h e t a n g e n t P T a t P . ∴ P Q ⊥ P T s i n c e t h e r a d i u s t h r o u g h t h e p o i n t o f c o n t a c t o f a t a n g e n t t o a c i r c l e i s p e r p e n d i c u l a r t o t h e t a n g e n t . ∴ ∠ Q P T = 90^{o} . A g a i n t h e m i n o r a r c P R s u b t e n d s ∠ P O R t o t h e c e n t r e O a n d ∠ P Q R t o t h e c i c u m f e r e n c e . ∴ ∠ P Q R = \frac{1}{2} \times ∠ P O R = \frac{1}{2} \times 72^{o} = 36^{o} . S o i n Δ P Q T ∠ Q T P = 180^{o} - (∠ P Q R + ∠ Q P T) (b y a n g l e s u m p r o p e r t y o f t r i a n g l e s) = 180^{o} - (36^{o} + 90^{o}) = 54^{o} . A n s - O p t i o n C .$

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In the given figure PQ is a diameter of a circle with centre O and PT is a tangent at P. QT meets the circle at R. If ∠POR=72o, ∠PTR is

In the given figure $P Q$ is a diameter of a circle with centre $O$ and $P T$ is a tangent at $P$ . $Q T$ meets the circle at $R$ . If $∠ P O R = 72^{o}$ , $∠ P T R$ is