The tangent at a point C of a circle and a diameter AB when extended intersect at P. If $∠ P C A = 110^{0}$ , find CBA [see Fig. 9.21].

50^{0}

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60^{0}

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70^{0}

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80^{0}

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Solution

The correct option is C $70^{0}$
$G i v e n - P C i s t h e t a n g e n t a t a p o i n t C o f t h e c i r c l e w i t h d i a m e t e r A B a n d c e n t r e O . A B, e x t e n d e d, m e e t s P C a t P & ∠ P C A = 110^{o} . T o f i n d o u t - ∠ C B A = C o n s t r u c t i o n - W e j o i n O C . S o l u t i o n - P C i s t h e t a n g e n t a t C a n d O C i s t h e r a d i u s f r o m O t o C . ∴ ∠ P C O = 90^{o} i . e ∠ O C A = 110^{o} - 90^{o} = 20^{o} . . . . . . . (i) N o w i n Δ O C A w e h a v e O C = O A (r a d i i o f t h e s a m e c i r c l e) ∴ Δ O C A i s i s o s c e l e s . ⟹ ∠ O C A = ∠ O A C o r ∠ B A C = 20^{o} . . . (i i) (f r o m i) A g a i n ∠ A C B i s t h e a n g l e a t t h e c i r c u m f e r e n c e s u b t e n d e d b y t h e d i a m e t e r A B a t C . S o ∠ A C B = 90^{o} . . . . . (i i i) ∠ C B A = 180^{o} - (∠ A C B + ∠ B A C) (a n g l e s u m p r o p e r t y o f t r i a n g l e s) = 180^{o} - (∠ A C B + ∠ B A C) = 180^{o} - 90^{o} - 20^{o} = 70^{o} A n s - 70^{o}$

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In the figure, tangent at a point C of a circle and a diameter AB when extended intersect at P. If $∠ P C A = 110^{\circ}$ , find $∠ C B A$

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