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Angle Subtended by an Arc of a Circle at the Centre

In the figure...

Question

In the figure, if OP||RS, $∠ O P Q = 110^{0} a n d ∠ Q R S = 130^{0}$ , then $∠ P Q R$ is equal to

40^{0}

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50^{0}

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60^{0}

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70^{0}

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Solution

The correct option is B $60^{0}$
As OP||RS,
$∠$ PQR $= 180 - (180 - 110) - (180 - 130)$ = $60$

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Similar questions

O P ∥ R S, ∠ O P Q = 110^{o}

∠ Q R S = 130^{o}

∠ P Q R

In the figure given below, if OP || RS, ∠OPQ = 110^o and ∠QRS = 130^o, then ∠PQR is equal to

In the figure given below, if OP || RS, ∠OPQ=110^o and ∠QRS=130^o, then ∠PQR is equal to

O P | | R S, ∠ O P Q = 110^{\circ} a n d ∠ Q R S = 130^{\circ}, t h e n ∠ P Q R

40^{\circ}

50^{\circ}

60^{\circ}

70^{\circ}

O P | | R S, ∠ O P Q = 110^{\circ} a n d ∠ Q R S = 130^{\circ}, t h e n ∠ P Q R

40^{\circ}

50^{\circ}

60^{\circ}

70^{\circ}

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