Given:
In ΔJKL, k = 4.1 cm, j = 3.8 cm and ∠J=63°.
To find:
The value of ∠K to the nearest 10th of a degree.
Solution:
According to the Law of Sine:
[tex]\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}[/tex]
We have, In ΔJKL, k = 4.1 cm, j = 3.8 cm and ∠J=63°.
Using the Law of Sine, we get
[tex]\dfrac{j}{\sin J}=\dfrac{k}{\sin K}[/tex]
[tex]\dfrac{3.8}{\sin (63^\circ)}=\dfrac{4.1}{\sin K}[/tex]
[tex]3.8\sin K=4.1\sin (63^\circ)[/tex]
[tex]\sin K=\dfrac{4.1\sin (63^\circ)}{3.8}[/tex]
[tex]\sin K=0.961349[/tex]
[tex]K=\sin^{-1}(0.961349)[/tex]
[tex]K=74.0182[/tex]
[tex]K\approx 74.0[/tex]
Therefore, the measure of angle K is 74.0 degrees.
