Answer:
[tex]x=6[/tex]
Step-by-step explanation:
we know that
If SQ is an angle bisector
then
[tex]m\angle TSQ=m\angle QSR[/tex]
we have
[tex]m\angle TSQ=(5x+3)^o[/tex]
[tex]m\angle OSR=(10x-27)^o[/tex]
substitute the given values
[tex](5x+3)^o=(10x-27)^o[/tex]
solve for x
Group terms
[tex]10x-5x=3+27[/tex]
Combine like terms
[tex]5x=30[/tex]
Divide by 5 both sides
[tex]x=6[/tex]
