Question

PQ is a diameter of circle and PR is chord such that angle RPQ equal to 30 degree the tangent at R intersect PQ produce at S prove that are equal RQ equal to QS​

Answer 1

Answer with Step-by-step explanation:

Let O be the center of circle

PQ is the diameter of circle

OP=QO=Radius of circle

In triangle OPR

OR=OP

Angle QPR=30 degrees

Angle RPO=Angle ORP=30 degrees

Angle made by two equal sides are equal

Angle subtended by half circle at the circumference of circle=90 degrees

Angle QRP=90 degrees

Angle ORQ=Angle QRP-angle ORP=90-30=60 degrees

Angle ORS=90 degrees

Radius is always perpendicular to tangent.

Angle QRS=Angle ORS-angle ORQ=90-60=30 degrees

$\angle PRS=\angle PRQ+\angle QRS=90+30=120^{\circ}$

In triangle PRS

$\angle PRS+\angle RPS+\angle PSR=180^{\circ}$

By triangle angles sum property

Substitute the values

$120+30+\angle PSR=180$

$150+\angle PSR=180$

$\angle PSR=180-150=30^{\circ}$

In triangle QSR

Angle QSR=Angle QRS=30 degrees

Therefore, QR=QS

Two sides which make equal angles are equal.

Hence, proved.

#Learns more:

https://brainly.in/question/1134715:Answered by Akash Mandal

Answer 2

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