Question

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

Answer 1

Step-by-step 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,

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