In the given figure, PQ is the diameter. $∠ S Q P$ is equal to

40^{\circ}

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30^{\circ}

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60^{\circ}

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50^{\circ}

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Solution

The correct option is C $60^{\circ}$
In the given figure PQRS is a cyclic quadrilateral, and opposite angles of a cyclic quadrilateral are supplementary so $∠ R$ + $∠ P$ will be $180$
$∠ S$ = $90$ (Angle in a semi circle is 90 degree)
In $△ P S Q$
$∠ P S Q + ∠ S Q P + ∠ S P Q = 180$
$90 + ∠ S Q P + 30 = 180$
$∠ S Q P = 180 - 120 = 60^{0}$

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In the given figure, PQ is the diameter. ∠SQP is equal to

The correct option is C 60∘In the given figure PQRS is a cyclic quadrilateral, and opposite angles of a cyclic quadrilateral are supplementary so ∠R + ∠P will be 180∠S=90(Angle in a semi circle is 90 degree)In △PSQ ∠PSQ+∠SQP+∠SPQ=18090+∠SQP+30=180∠SQP=180−120=600

In the given figure, PQ is the diameter. $∠ S Q P$ is equal to