Using the law of cosine for Triangle KJL, we can write:
[tex] j^{2} = k^{2} + l^{2}-2(k)(l)cos(J) \\ \\
2(k)(l)cos(J)=k^{2} + l^{2}- j^{2} \\ \\
cos(J)= \frac{k^{2} + l^{2}- j^{2}}{2(k)(l)} [/tex]
Using the values of k,j and l, we can write:
[tex]cos(J)= \frac{ 4^{2} + (2.89)^{2} - 3^{2} }{2(4)(2.89)} \\ \\
cos(J)= 0.664 \\ \\
J= cos^{-1}(0.664) \\ \\
J=48.39 [/tex]
Rounding to nearest integer, the measure of angle J will be 48 degrees.
So option B gives the correct answer
