Question

In a cyclic quadrilateral PQRS, angle QPR=40 and angle PSQ=70 , then find angle PQR ?​

Answer 1

Step-by-step explanation:

PQRS is a cyclic quadrilateral.

Then, ∠PQR+∠PSR=180∘ ...[Opposite angles of cyclic quadrilateral are supplementary]

⟹ ∠PQR+150o=180∘

⟹ ∠PQR=180∘−150∘=30∘.

In △PQR,

∠PRQ=90∘ (Angle of a semicircle)

Then, ∠RPQ+90∘+30∘=180∘ ...[Angle sum property]

⇒∠RPQ+120∘=180∘

⇒∠RPQ=60

Answer 2

• ∠PQR is equal to 70° .

Given :- In a cyclic quadrilateral PQRS,

• ∠QPR = 40°
• ∠PSQ = 70°

To Find :-

• ∠PQR = ?

Concept used :-

• A chord subtends equal angles at any part of the circumference of the circle .
• Opposite angles of cyclic quadrilateral are supplementary .

Solution :-

from image we can see that,

→ ∠QPR = 40° { given }

now,

→ ∠QPR = ∠QSR { chord QR subtends equal angle at the circumference of the circle }

So,

→ ∠QSR = 40° ------- Equation (1)

also,

→ ∠PSQ = 70° { given }

adding Equation (1),

→ ∠PSQ + ∠QSR = 70° + 40°

→ ∠PSR = 110° ---------- Equation (2)

now,

→ ∠PSR + ∠PQR = 180° { Opposite angles of cyclic quadrilateral are supplementary }

→ 110° + ∠PQR = 180°

→ ∠PQR = 180° - 110°

→ ∠PQR = 70° (Ans.)

Hence, ∠PQR is equal to 70° .

Learn more :-

In the figure along side, BP and CP are the angular bisectors of the exterior angles BCD and CBE of triangle ABC. Prove ∠BOC = 90° - (1/2)∠A .

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