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

User Bartzilla
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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)}

In triangle JKL j=9, k=5 and angle L =43 degrees. find angle J-example-1
User Laz
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