In a convex quadrilateral PQRS, SPR = 80°

QPR = 40°, SQR = 40° and PR = PS. If

diagonals PR and QS intersect at O, then the

measure of POQ is

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kindly provide complete and detailed solution , required urgently.
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 Lets consider a random quadrilateral having the given angles and properties.
Given: <SPR=80, <SQR=40, <RPQ=40 & PR=PS
To Find: <POQ
In Tr PRS as PR=PS & <SPR=80, <PSR=<PRS=50.    - Angle some property isosceles in triangle PRS.

Lets consider a circle with center P and radius PR.
It is obvious that S lies on the circle as PR=PS.

If we take any point on the major arc, the angle subtended by the chord SR at that point will be half the angle subtended by it at the center.  So Q can be visualised as any point on the major arc so that <SQR=40 as given. 

Lets draw a line from P which makes an angle 40 with PR and lets call the point where the line meets the circle as Q. 
If we now join QR we get a quadrilateral that satisfies every condition given in the question. Hence we can consider this quadrilateral to solve. 
Now in Tr PQR, PQ=PR=radius of the circle. Hence < PQR= <QRP=(180 - <QPR) / 2 = 70
Therefore <PQO = 70 - 40 = 30
and hence <POQ = 180 - (40 + 30) = 110.
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110 degree
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