Question

PQ AND PR are tangents to a circle wirh with centre A if angle QPA = 27 find angle QAR

Answer 1

Here

Given

<QPA= 27*

Proof: In triangle QPA and RPA

QA=AR............(radii of circle)

PQ=PR............(tangent from a

point to the circle

Circle are equal)

PA=PA............(common side)

:. Triangle QPA congruent to RPA

by S-S-S

Hence <QPA=<RPA

:.<QPR=27+27

<QPR=54

Hence using property tangent is perpendicular to radius

:. <PQA=<PRA=90*

:. <PQA+<PRA+<QPR+<QAR=360*..........(sum of quadilateral)

:.90*+90*+54*+<QAR=360*

:.234*+<QAR=360*

:.<QAR=360-234

hence <QAR=126*

Answer 2

Here

Given

<QPA= 27*

Proof: In triangle QPA and RPA

QA=AR............(radii of circle)

PQ=PR............(tangent from a

point to the circle

Circle are equal)

PA=PA............(common side)

:. Triangle QPA congruent to RPA

by S-S-S

Hence <QPA=<RPA

:.<QPR=27+27

<QPR=54

Hence using property tangent is perpendicular to radius

:. <PQA=<PRA=90*

:. <PQA+<PRA+<QPR+<QAR=360*..........(sum of quadilateral)

:.90*+90*+54*+<QAR=360*

:.234*+<QAR=360*

:.<QAR=360-234

hence <QAR=126*