Step 1
Given;
[tex]12^2=(4.5)^2+10^2-2(4.5)(10)cosB[/tex]
Required; To find angle B
Step 2
[tex]\begin{gathered} 4.5^2+10^2-90\cos\left(B\right)=12^2 \\ 20.25+100-90\cos\left(B\right)=12^2 \\ -90\cos\lparen B)+120.25=12^2 \\ -90\cos\left(B\right)+120.25=144 \\ \\ \end{gathered}[/tex][tex]\begin{gathered} -90\cos\left(B\right)=23.75 \\ \frac{-90\cos\left(B\right)}{-90}=\frac{23.75}{-90} \\ cos(B)=-\frac{19}{72} \\ B=\cos^{-1}(-\frac{19}{72}) \\ B=105.3009409^{\circ} \\ B\approx105^{\circ} \end{gathered}[/tex]
Answer;
[tex]B=105^{\circ}\text{ to the nearest degree}[/tex]
