PQ and QR are tangents to the circle with centre O, at P and R respectively. Find value of x where angle PQR is 50 degree.
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Answered by
9
Answer:
∠POR=180−50=130
0
∠PSR=
2
130
=65
0
Join 'S' and 'O', then we get
∠SRO=∠OSR=20
0
[since OR=OS=radius]
∠OSP=65−20=45
0
∠SPO=45
o
=∠OSP
[since OS=OP=radius]
∠OPT=90
0
[OP⊥TQ]
x
0
=∠SPT=90
0
−45
0
∴x
0
=45
0
#HAVE A NICE DAY
Answered by
5
∠POR=180−50=130
0
∠PSR=
2
130
=65
0
Join 'S' and 'O', then we get
∠SRO=∠OSR=20
0
[since OR=OS=radius]
∠OSP=65−20=45
0
∠SPO=45
o
=∠OSP
[since OS=OP=radius]
∠OPT=90
0
[OP⊥TQ]
x
0
=∠SPT=90
0
−45
0
∴x
0
=45
0
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