We will use the following trigonometric laws, sine law and cosine law.
Cosine law is :
[tex]c^2=a^2+b^2-2ab\cos C[/tex]
Sine law is :
[tex]\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}[/tex]
From the problem, we have B = 74 degrees, a = 38 and c = 41
We need first to find the value of b using cosine law, note that the missing value is b, so we will rewrite the cosine law as b in terms of a and c :
[tex]b^2=a^2+c^2-2ac\cos B[/tex]
Substitute the given values and solve for b :
[tex]\begin{gathered} b^2=38^2+41^2-2(38)(41)\cos 74 \\ b^2=3125-3116\cos 74 \\ b=\sqrt[]{3125-3116\cos 74} \\ b=47.60 \end{gathered}[/tex]
Next is to use the sine law with B = 74 degrees, b = 47.60 and a = 38.
[tex]\begin{gathered} \frac{b}{\sin B}=\frac{a}{\sin A} \\ \frac{47.60}{\sin 74}=\frac{38}{\sin A} \\ \sin A=\frac{38\sin 74}{47.60} \\ \sin A=0.7674 \\ \arcsin (\sin A)=\arcsin (0.7674) \\ A=50.12 \end{gathered}[/tex]
The answer is C. m∠A = 50.1 degrees
