Question

47.
In the diagram, PQ and QR are tangents to the circle centre O, at P and R
respectively. Find the value of x
R​

Answer 1

Answer:

55

Step-by-step explanation:

∠POR=180−50=130  

0

 

∠PSR=  

2

130

​  

=65  

0

 

Join 'S' and 'O', then we get

∠SRO=∠OSR=30  

0

 

[since OR=OS=radius]

∠OSP=65−30=35  

0

 

∠SPO=35  

o

=∠OSP

[since OS=OP=radius]

∠OPT=90  

0

[OP⊥TQ]

x  

0

=∠SPT=90  

0

−35  

0

 

∴x  

0

=55  

0