In the adjoining figure- if - QP R -67- and - SPR -72- then - QRS is equal toA- 41B- 23C- 67D- 18

The correct option is A 41 Clearly, it is a cyclic quadrilateral. So, ∠ QRS = 180^∘ ∠ QPR ∠ SPR [Sum of opposite angles of a cyclic quadrilateral is 18 ...

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Byju's Answer

Standard X

Mathematics

Sum of Opposite Angles of a Cyclic Quadrilateral

In the adjoin...

Question

In the adjoining figure, if $∠ Q P R = 67^{\circ}$ and $∠ S P R = 72^{\circ}$ , then $∠ Q R S$ is equal to

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Solution

The correct option is A
41

Clearly, it is a cyclic quadrilateral.
So, $∠ Q R S = 180^{\circ} - ∠ Q P R - ∠ S P R$ [Sum of opposite angles of a cyclic quadrilateral is $180^{\circ}$ ]

$\Rightarrow 180^{\circ} - (67^{\circ} + 72^{\circ}) = 41^{\circ}$

$∴ ∠ Q R S = 41^{\circ}$

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Similar questions

In the adjoining figure, if $∠ Q P R = 60^{\circ}$ and $∠ S P R = 70^{\circ}$ , then $∠ Q R S$ = ___.

Q. In the figure, if

∠ Q P R = 67^{0}

and

∠ S P R = 72^{0}

and

R P

is a diameter of the circle, then

∠ Q R S

is equal to:

PQRS is a cyclic quadrilateral such that PR is a diameter of the circle. If $∠ Q P R = 67^{o} a n d ∠ S P R = 72^{o}$ , then $∠ Q R S =$

Q. In the figure, PQ is the diameter of the semicircle in which

∠ S P R = 30^{\circ}; ∠ Q P R = 20^{\circ}

;
then

∠ S R P

equals :

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is cyclic quadrilateral such that

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is the diameter of the circle.

∠ Q P R = 50^{\circ}, ∠ S P R = 66^{\circ},

then

∠ Q R S

is:

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In the adjoining figure, if ∠QPR=67∘ and ∠SPR=72∘, then ∠QRS is equal to

The correct option is A 41 Clearly, it is a cyclic quadrilateral. So, ∠QRS=180∘−∠QPR−∠SPR [Sum of opposite angles of a cyclic quadrilateral is 180∘] ⇒180∘−(67∘+72∘)=41∘ ∴∠QRS=41∘

In the adjoining figure, if $∠ Q P R = 67^{\circ}$ and $∠ S P R = 72^{\circ}$ , then $∠ Q R S$ is equal to

The correct option is A
41

Clearly, it is a cyclic quadrilateral.
So, $∠ Q R S = 180^{\circ} - ∠ Q P R - ∠ S P R$ [Sum of opposite angles of a cyclic quadrilateral is $180^{\circ}$ ]

$\Rightarrow 180^{\circ} - (67^{\circ} + 72^{\circ}) = 41^{\circ}$

$∴ ∠ Q R S = 41^{\circ}$