In the isosceles triangle PQR- the vertical - P -50- The circle passing through Q and R - intersects PQ in S and PR in T- ST is joined- - PST in degrees is

The correct option is B 65 #160;Given Δ PQR is an isosceles one. ∠ QPR= 50 ^ o . A circle, passing through Q & R, i ...

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Standard VIII

Mathematics

Angle Sum Property of a Quadrilateral

In the isosce...

Question

In the isosceles triangle PQR, the vertical $∠ P = 50^{\circ}$ . The circle passing through Q and R , intersects PQ in S and PR in T. ST is joined. $∠$ PST in degrees is

115

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135

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100

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Solution

The correct option is B 65
$G i v e n - Δ P Q R i s a n i s o s c e l e s o n e . ∠ Q P R = 50^{o} . A c i r c l e, p a s s i n g t h r o u g h Q & R, i n t e r s e c t s P Q a t S a n d P R a t T . S & T a r e j o i n e d . T o f i n d o u t - ∠ P S T = ? S o l u t i o n - Δ P Q R i s a n i s o s c e l e s o n e . ∴ ∠ P Q R = ∠ P R Q ⟹ ∠ P Q R + ∠ P R Q = 2 (∠ P Q R O R ∠ P R Q) ∴ ∠ Q P R + ∠ P Q R + ∠ P R Q = ∠ Q P R + 2 ∠ P Q R = 50^{o} + 2 ∠ P Q R = 180^{o} . (A n g l e s u m p r o p e r t y o f t r i a n g l e s) . i . e ∠ P Q R = 65^{o} = ∠ P Q R . N o w Q S T R i s a c y c l i c q u a d r i l a t e r a l . ∴ ∠ S Q R + ∠ S T R = 180^{o} a n d ∠ T R Q + T S Q = 180^{o} s i n c e t h e s u m o f t h e o p p o s i t e a n g l e s o f a c y c l i c q u a d r i l a t e r a l i s 180^{o} . S o ∠ T S Q = 180^{o} - ∠ T R Q = 180^{o} - 65^{o} = 115^{o} . i . e ∠ P S T = 180^{o} - 115^{o} = 65^{o} . (l i n e a r p a i r)$

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In the isosceles triangle PQR, the vertical ∠P=50∘. The circle passing through Q and R , intersects PQ in S and PR in T. ST is joined. ∠ PST in degrees is

In the isosceles triangle PQR, the vertical $∠ P = 50^{\circ}$ . The circle passing through Q and R , intersects PQ in S and PR in T. ST is joined. $∠$ PST in degrees is