Question

In triangle JKL j=9, k=5 and angle L =43 degrees. find angle J

Answer 1

Explanation:

By the Law of Cosines,

`l^(2)=5^(2)+9^(2)-2(5)(9)(\cos 43^(\circ))\\l^(2)=106-90 \cos 43^(\circ)\\l=\sqrt{106-90 \cos 43^(\circ)`

This means that by the Law of Sines,

`(\sin J)/(9)=\frac{\sin 43^(\circ)}{\sqrt{106-90 \cos 43^(\circ)}}\\\sin J=9 \left(\frac{\sin 43^(\circ)}{\sqrt{106-90 \cos 43^(\circ)}} \right)\\J=\sin^(-1) \left(9\left(\frac{\sin 43^(\circ)}{\sqrt{106-90 \cos 43^(\circ)}} \right) \right)\\J \approx \boxed{75.55^(\circ)}`