Notice that angles BDA and CDA lay in the same line
Therefore,
[tex]\begin{gathered} m\angle BDA+m\angle CDA=180 \\ \rightarrow98+m\angle CDA=180 \\ \rightarrow m\angle CDA=180-98 \\ \Rightarrow m\angle CDA=82 \end{gathered}[/tex]
Now, since segment AD bisects angle A, we know that angles BAD and DAC are the same, and that
[tex]m\angle A=2\cdot m\angle DAC[/tex]
Remember that the sum of the interior angles of a triangle is 180°. This way,
[tex]\begin{gathered} m\angle DCA+m\angle CDA+m\angle DAC=180 \\ \rightarrow69+82+m\angle DAC=180 \\ \rightarrow151+m\angle DAC=180 \\ \rightarrow m\angle DAC=180-151 \\ \Rightarrow m\angle DAC=29 \end{gathered}[/tex]
Thereby,
[tex]\begin{gathered} m\angle A=m\angle DAC \\ \rightarrow m\angle A=2\cdot29 \\ \Rightarrow m\angle A=58 \end{gathered}[/tex]
Using the fact that the sum of the interior angles of a triangle is 180°, we can state that
[tex]\begin{gathered} m\angle A+m\angle B+m\angle C=180 \\ \rightarrow58+m\angle B+69=180 \\ \rightarrow m\angle B+127=180 \\ \rightarrow m\angle B=180-127 \\ \Rightarrow m\angle B=53 \end{gathered}[/tex]
Thereby, we can conclude that angle B measures 53°
