Question

In ΔJKL, k = 4.1 cm, j = 3.8 cm and ∠J=63°. Find all possible values of ∠K, to the nearest 10th of a degree.

Answer 1

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:

$\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}$

We have, In ΔJKL, k = 4.1 cm, j = 3.8 cm and ∠J=63°.

Using the Law of Sine, we get

$\dfrac{j}{\sin J}=\dfrac{k}{\sin K}$

$\dfrac{3.8}{\sin (63^\circ)}=\dfrac{4.1}{\sin K}$

$3.8\sin K=4.1\sin (63^\circ)$

$\sin K=\dfrac{4.1\sin (63^\circ)}{3.8}$

$\sin K=0.961349$

$K=\sin^{-1}(0.961349)$

$K=74.0182$

$K\approx 74.0$

Therefore, the measure of angle K is 74.0 degrees.