if pr and rq are tangent of circle with centre o such that angle rpq=50 deg ,find angle poq
Answers
Answer:
It is given that, ∠RPQ=30
o
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]
∴ ∠PRQ=∠PQR [Angles opposite to equal sides are equal in a triangle. ]
In △PQR,
∠RQP+∠QRP+∠RPQ=180
o
[Angle sum property of a triangle ]
⇒ 2∠RQP+30
o
=180
o
⇒ 2∠RQP=150
o
⇒ ∠RQP=75
o
so ∠RQP=∠QRP=75
o
⇒ ∠RQP=∠RSQ=75
o
[ By Alternate Segment Theorem]
Given, RS∥PQ
∴ ∠RQP=∠SRQ=75
o
[Alternate angles]
⇒ ∠RSQ=∠SRQ=75
o
∴ QRS is also an isosceles triangle. [Since sides opposite to equal angles of a triangle are equal.]
⇒ ∠RSQ+∠SRQ+∠RQS=180
o
[Angle sum property of a triangle]
⇒ 75
o
+75
o
+∠RQS=180
o
⇒ 150
o
+∠RQS=180
o
∴ ∠RQS=30
o
Step-by-step explanation:
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