To solve this question, we would use cosine rule which is given as
[tex]a^2=b^2+c^2-2bc\cos A[/tex]
Our values have been defined for us and we will proceed to evaluate
[tex]\begin{gathered} a^2=4^2+10^2-2(4)(10)\cos 41 \\ a^2=16+100-80\cos 41 \\ a^2=116-60.376 \\ a^2=55.624 \\ \text{take the square root of both sides} \\ a=\sqrt[]{55.624} \\ a=7.458\approx7.5 \end{gathered}[/tex]
From the calculations above, the value of the missing side a is 7.5 units
To find angle B,
we can use sine rule
[tex]\begin{gathered} \frac{a}{\sin A}=\frac{b}{\sin B} \\ \frac{7.5}{\sin 41}=\frac{4}{\sin B} \\ \sin B=\frac{4\times\sin 41}{7.5} \\ \sin B=0.3498 \\ B=\sin ^{-1}0.3498 \\ B=20.5^0 \end{gathered}[/tex]
We can still approach C with sine rule or sum of angle in a triangle
[tex]\begin{gathered} A+B+C=180 \\ 41+20.5+C=180 \\ c=118.5^0 \end{gathered}[/tex]
From the calculations above, the value of a = 7.5 , B = 22⁰ and C = 118.5⁰ respectively which is option B
