PQRST is a pentagon inside a circle with centre O such that PQ=QR=RS and angle PQR=128 RO and - Maths - Quadrilaterals

Question

PQRST is a pentagon inside a circle with centre
O such that PQ=QR=RS and angle
PQR=128 RO and
SO are joined find angle PTQ ,PTS,
ROS

Utsav · Accepted Answer

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RO ,SO,TQ,PR QS is joined

Now in∆PQR,we have PQ=QR hence ∠QPR=∠QRP Also
∠PQR+∠QPR+∠QRP = 180°(sum of interior ∠s of a
triangle)∠PQR =
128°(given)∴∠QPR=∠QRP=180°-128°2=26°Now
∠QSR and ∠QPR are the angles subtended by the same chord QR
or say the angles in the same segment∴ ∠QSR= ∠QPR
=26°Now in∆QRS , since QR=RS hence ∠QRS=∠SQR
=26°Also ∠QRS = 180°-26°-26°=128°Now arc RS
subtend ∠ROS at the centre and ∠SQR at the circleAs we know
the angle subtended by an arc of a circle at the centre is double
the angle subtended by it at any point on the remaining part of the
circle∴∠ROS =2∠SQR =2×26°=52°Now in
cyclic quadrilateral QRST ,since sum of opposite angles is
180°Hence ∠QRS+∠QTS =180°⇒∠QTS
=180°-128°=52°Also from the figure we can say that
∠PQS+∠SQR =∠PQR⇒∠PQS =∠PQR-∠SQR
=128°-26°=102°Now in cyclic quadrilateral PQST ,since
sum of opposite angles is 180Hence
∠PQS+∠PTS=180°⇒∠PTS =180°-∠PQS
=180°-102°=78°Now from the figure we can say
that∠PTQ+∠QTS =∠PTS⇒∠PTQ=78°-∠QTS
=78°-52°=26°

PQRST is a pentagon inside a circle with centre O such that PQ=QR=RS and angle PQR=128 . RO and SO are joined find angle PTQ ,PTS, ROS.