Answer:
[tex]48.5654 \textdegree[/tex]
Step-by-step explanation:
Let [tex]D[/tex] be the mid point of [tex]AB[/tex]
Now in [tex]\Delta ACD\ and\ \Delta BCD[/tex]
[tex]AC=CB \ (given)\\CD=CD \ (common\ side)\\AD=DB \ (D\ is\ mid\ point\ of\ AB)[/tex]
[tex]Hence\ \Delta ACD\cong\Delta BCD[/tex]
[tex]\angle A=\angle B\\\angle ACD=\angle BCD\\\angle ADB=\angle BDC[/tex]
[tex]\angle ADB+\angle BDC=180\\2\angle ADB=180\\\angle ADB=90[/tex]
[tex]in \Delta BCD\\\cos\angle B=\frac{BD}{BC}\\ =\frac{45}{2\times34}\\ =\frac{45}{68} \\\angle B=\cos^{-1}(\frac{45}{68} )\\\angleB=48.5654\textdegree[/tex]
