Question

in figure ∆PQR is such that PQ=PR and <QPR=50° find <PST.​

Answer 1

Answer:

65

Step-by-step explanation:

given:- pq=pr

angle pqr = angle prq

in pqr by angle addition property'

angle p +angle q + angle r

let angle r be x and so angle p will be x

and let x +angle x +angle 50 = 180

2x + 50 =180

2x= 130

x = 65

in pqr and pst

angle s = angle q corresponding angle

angle t = angle r. corresponding angle

by Angle Angle test pqr =pst

therefore angle pst=pqr and angle pts=angle prq

therefore angle pst and angle pts =65

Answer 2

Answer:

<PST = 65°

Step-by-step explanation:

1. since <QPR = 50°

and PR =PQ

therefore, ΔPQR is an isoscles Δ

by which, <PQR= <PRQ

2.we get, <QPR +< PQR +<PRQ =180°

= 50° + 2<PQR= 180.°

=<PQR=65°

3.now, as STRQ is cyclic quadrilateral

therefore, <PQR + <QST = 180°{as opp. angles of cylic quadrilateral made supplementary angles}

=<QST = 115°

now, by linear pair <PST= 65°