Answer:
[tex]m<QOB=75\°[/tex]
Step-by-step explanation:
we know that
The incenter of a triangle is the meeting point of the three angle bisectors of a triangle.
so
[tex]m<QBO=m<OBR[/tex]
we have
[tex]m<OBR=15\°[/tex] -----> given problem
so
[tex]m<QBO=15\°[/tex]
Remember that
In a triangle the sum of the internal angles must be equal to [tex]180\°[/tex]
so
In the right triangle OQB
[tex]m<QBO+m<QOB+90\°=180\°[/tex]
substitute
[tex]15\°+m<QOB+90\°=180\°[/tex]
[tex]m<QOB=180\°-90\°-15\°=75\°[/tex]
