Tangents PQ and PR are drawn to a circle such that angle RPQ= 30 degree
A chord RS is drawn parallel to tangent PQ.
Find angle RQS...??? ANSFAST...
tangents PQand PR are drawn to a circles such that angles RPQ=30degree a chord RS is drawn parallel to tangents PQ find angle RQS
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rshimpy86... Can u explain this in detail...
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Tangents drawn from an external point to a circle are equal
Hence, PQ = PR
And PQR is an isosceles triangle
ThereforeRQP =QRP
RQP +QRP +RPQ = 180[Angles in a triangle]
2RQP + 30 = 180
2RQP = 150
RQP =QRP = 75
RQP =RSQ= 75 [ Angles in alternate Segment Theoremstates that angle between chord and tangent is equal to the angle in the alternate segment]
RS II PQ
Therefore RQP = SRQ= 75 [They are alternate angles]
RSQ = SRQ = 75
Therefore QRS is also an isosceles triangle
RSQ + SRQ + RQS = 180 [Angles in a triangle]
75 + 75 + RQS = 180
150 + RQS = 180
RQS = 30
Required Answer: RQS = 30
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in triangle PQR
RPQ= 30
let consider PRQ AND RQP=x
(PQ=PR)
PRQ+QPR+RQP=180 (ASP)
30 + 2x = 180
x= 75
OPR=90 (PR =TANGENT)
ORP=90 and QRP = 75
ORQ=15
SR ll PQ
then,SRQ =RQP
RQP= 75(proved above)=SRQ
SRQ= SRO + ORQ
75= SRO+15
SRO=60= OSR( OR=OS radius of circle)
in triangle OSR
OSR+SRO+ROS=180(ASP)
SOR=60
2 TIMES SOR = SQR
RQS =60/2
=30
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RPQ= 30
let consider PRQ AND RQP=x
(PQ=PR)
PRQ+QPR+RQP=180 (ASP)
30 + 2x = 180
x= 75
OPR=90 (PR =TANGENT)
ORP=90 and QRP = 75
ORQ=15
SR ll PQ
then,SRQ =RQP
RQP= 75(proved above)=SRQ
SRQ= SRO + ORQ
75= SRO+15
SRO=60= OSR( OR=OS radius of circle)
in triangle OSR
OSR+SRO+ROS=180(ASP)
SOR=60
2 TIMES SOR = SQR
RQS =60/2
=30
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