Construct a circle with radius equal to $3 c m$ . Draw two tangents to it inclined at an angle of $60$ at their point of intersection. Measure their lengths and verify the results by calculation.

B y m e a s u r e m e n t : 120^{o} . B y c a l c u l a t i o n : 60^{o}

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B y m e a s u r e m e n t : 60^{o} . B y c a l c u l a t i o n : 120^{o}

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B y m e a s u r e m e n t : 60^{o} . B y c a l c u l a t i o n : 60^{o}

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B y m e a s u r e m e n t : 120^{o} . B y c a l c u l a t i o n : 120^{o}

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Solution

The correct option is C $B y m e a s u r e m e n t : 60^{o} . B y c a l c u l a t i o n : 60^{o}$
$O i s t h e c e n t r e o f a c i r c l e o f r a d i u s 3 c m . T o c o n s t r u c t - t w o t a n g e n t s P A & P B t o t h e c i r c l e s u c h t h a t ∠ A P B = 60^{o} . C o n s t r u c t i o n - (I) A c i r c l e o f r a d i u s 3 c m w i t h c e n t r e O i s d r a w n . (I I) O n a n y r a d i u s O A a n a n g l e O A B = 30^{o} i s c o n s t r u c t e d . A B m e e t s t h e c i r c u m f e r e n c e a t B . (I I I) W e j o i n O B . (I V) A t A & B w e c o n s t r u c t t w o p e r p e n d i c u l a r s A P & B P, s u c h t h a t O A ⊥ A P & O B ⊥ B P . A P & B P i n t e r s e c t a t P . T h e m e a s u r e m e n t o f ∠ A B P = 60^{o} . T h e n P A & P B a r e t h e r e q u i r e d t a n g e n t s t o t h e g i v e n c i r c l e . J u s t i f i c a t i o n - B y c o n s t r u c t i o n, O A ⊥ A P & O B ⊥ B P a n d O A & O B a r e r a d i i . ∴ P A & P B a r e t a n g e n t s t o t h e g i v e n c i r c l e f r o m P a t A & B r e s p e c t i v e l y . (s i n c e t h e r a d i u s t h r o u g h t h e p o i n t o f c o n t a c t o f a t a n g e n t t o a c i r c l e i s p e r p e n d i c u l a r t o t h e t a n g e n t .) I n Δ O A B O A = O B (r a d i i o f t h e s a m e c i r c l e . ∴ Δ O A B i s i s o s c e l e s w i t h A B a s b a s e . i . e ∠ O A B = ∠ O B A = 30^{o} . N o w O A ⊥ A P & O B ⊥ B P ⟹ ∠ O B P = ∠ O A P = 90^{o} . ∴ ∠ A B P = ∠ O B P - ∠ O B A = 90^{o} - 30^{o} = 60^{o} a n d ∠ B A P = ∠ O A P - ∠ O A B = 90^{o} - 30^{o} = 60^{o} . S o, i n Δ A P B ∠ A P B = 180^{o} - (∠ A B P + ∠ B A P) = 180^{o} - (60^{o} + 60^{o}) = 60^{o} . (b y a n g l e s u m p r o p e r t y o f t r i a n g l e s) . S o ∠ A P B = 60^{o} b y m e a s u r e m e n t a s w e l l a s b y r e a s o n . A n s - O p t i o n C$

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