# Tangents PQ and PR are drawn to a circle such that angle RPQ= 30 degreeA 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

• -18

rshimpy86... Can u explain this in detail...

• -13

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

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i too found this sum puzzling......hope my answer helps

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As the SEGMENT THEOREM is not in many revised NCERT books so here is an alternative method , Construct line through Q such that it is perpendicular to QP and name the piont of contact on line SR as A.Then,AQR=90°But we found that RQP =75°so, AQR= AQP-RQPAQR= 90-75=15°Therefore,SQR=2AQR= 30° !Hope you find my answer helpful ! :)
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@MeetDholaria - thnx a lot .. But I think u meant AQP = 90 instead of AQR =90 ..
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i dont know
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yaaro ye kaise hoga jldi solution bta do 10 march ko mera exam h maths ka
• -14
yes the above answer is correct
• -14

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
• 83
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
• -1
RPQ= 90
PR     = PQ
RQP    = <PRQ
2RQP + 30  = 180
2RQP   = 150
RQP    = 75
RSQ  = 75 (Alternate Segment Theorem)
RS ll PQ and SQ be the transversal
SQP = 180 - 75
= 105
RQS = 105 - <RQP
= 30
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thanks
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PQ = PR Since tangents drawn from an external point to a circle are equal. And PQR is an isosceles triangle thus, ∠RQP = ∠QRP ∠RQP + ∠QRP + ∠RPQ = 180° [Angle sum property of a triangle] 2∠RQP + 30° = 180° 2∠RQP = 150° ∠RQP = ∠QRP = 75° ∠RQP = ∠RSQ = 75° [ Angles in alternate Segment Theorem states that angle between chord and tangent is equal to the angle in the alternate segment] RS is parallel to PQ Therefore ∠RQP = ∠SRQ = 75° [Alternate angles] ∠RSQ = ∠SRQ = 75° Therefore QRS is also an isosceles triangle ∠RSQ + ∠SRQ + ∠RQS = 180° [Angle sum property of a triangle] 75° + 75° + ∠RQS = 180° 150° + ∠RQS = 180° ∠RQS = 30°
• -21
PQ = PR Since tangents drawn from an external point to a circle are equal. And PQR is an isosceles triangle thus, ∠RQP = ∠QRP ∠RQP + ∠QRP + ∠RPQ = 180° [Angle sum property of a triangle] 2∠RQP + 30° = 180° 2∠RQP = 150° ∠RQP = ∠QRP = 75° ∠RQP = ∠RSQ = 75° [ Angles in alternate Segment Theorem states that angle between chord and tangent is equal to the angle in the alternate segment] RS is parallel to PQ Therefore ∠RQP = ∠SRQ = 75° [Alternate angles] ∠RSQ = ∠SRQ = 75° Therefore QRS is also an isosceles triangle ∠RSQ + ∠SRQ + ∠RQS = 180° [Angle sum property of a triangle] 75° + 75° + ∠RQS = 180° 150° + ∠RQS = 180° ∠RQS = 30°
• 14
Mhmmm, the question was asked on 2012 and first answer came in like 2014
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I know
• -12
Actually In Rd,answer of this question is given 75*......so which answer we sgould consider
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it is same as the other answers
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Type your question student-name Aneesa Attaullah asked in Math 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... Share 30 Follow 20 ANSWER NOW student-name Lakshita Malhotra answered this 69 helpful votes in Math, Class XI-Science 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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Sol: Given ∠RPQ=30° and PR and PQ are tangents drawn from P to the same circle. Hence PR = PQ [Since tangents drawn from an external point to a circle are equal in length] Therefore, ∠PRQ = ∠PQR [Angles opposite to equal sides are equal in a triangle] In ΔPQR ∠RQP + ∠QRP + ∠RPQ = 180° [Angle sum property of a triangle] 2∠RQP + 30° = 180° 2∠RQP = 150° ∠RQP = 75° Hence, ∠RQP = ∠QRP = 75° ∠RQP = ∠RSQ = 75° [ By Alternate Segment Theorem] Given, RS || PQ Therefore ∠RQP = ∠SRQ = 75° [Alternate angles] ∠RSQ = ∠SRQ = 75° Therefore QRS is also an isosceles triangle. [Since sides opposite to equal angles of a triangle are equal.] ∠RSQ + ∠SRQ + ∠RQS = 180° [Angle sum property of a triangle] 75° + 75° + ∠RQS = 180° 150° + ∠RQS = 180° Therefore, ∠RQS = 30
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Answer of above question is following • -5
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