ΔPQR is equilateral and ΔQRS is a right angled triangle. Calculate the angles marked as x and y
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Step-by-step explanation:
Here, PQRS is a rectangle.
As we know in rectangle both the diagonals are equal.
⇒ PR=QS
Also diagonals bisect each other.
⇒ PO=QO
⇒ ∠OPQ=∠PQO [ Base angles of an equal sides are also equal ]
⇒ ∠OPQ=24
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[ Given ]
∴ ∠PQO=24
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In △PQO,
⇒ ∠OPQ+∠PQO+∠QOP=180
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⇒ 24
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+24
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+x=180
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⇒ 48
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+x=180
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∴ x=132
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Since, PQRS is a rectangle, PQ∥SR and PR is a transversal.
⇒ ∠QPR=∠SRP [ Alternate angles ]
therefore ∠SRP=24
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⇒ ∠SRP+∠PRQ=90
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[ Angle of an rectangle ]
⇒ 24
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+y=90
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∴ y=66
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