please help request.........please
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The answer for the question is 60°
Destroyer48:
so, x= 60°
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0
O is the center.
QR=OP
<ORP=20 degree
To find x.
OP=OQ →1 [radii of the circle with center O]
But QR=OP →2
From →1 and →2.
QR=OQ
So <PRT=<ROQ=20 degree
<OQP=<PRT+<ROQ {exterior angle property}
=20+20
<OQP=40 degrees
OP=OQ
So <OQP=<OPQ=40 degrees
<POQ=180 - <OQP - <OPQ {angle sum property}
=180-40-40
<POQ=100 degrees
POR is a line.
So <POR=180 degrees
<ROQ+<POQ+<x=180
20+100+x=180
x=180-20-100
x=60 degrees
QR=OP
<ORP=20 degree
To find x.
OP=OQ →1 [radii of the circle with center O]
But QR=OP →2
From →1 and →2.
QR=OQ
So <PRT=<ROQ=20 degree
<OQP=<PRT+<ROQ {exterior angle property}
=20+20
<OQP=40 degrees
OP=OQ
So <OQP=<OPQ=40 degrees
<POQ=180 - <OQP - <OPQ {angle sum property}
=180-40-40
<POQ=100 degrees
POR is a line.
So <POR=180 degrees
<ROQ+<POQ+<x=180
20+100+x=180
x=180-20-100
x=60 degrees
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