Answer:
[tex]K \approx 51.2^\circ[/tex]
[tex]K \approx 128.8^\circ[/tex]
Step-by-step explanation:
Given
[tex]k = 480[/tex]
[tex]j = 420[/tex]
[tex]\angle J = 43^\circ[/tex]
Required
Find the measure of K
The value of K ranges from: [tex]0^\circ \le K \le 180^\circ[/tex]
To do this, we apply sine rule;
[tex]\frac{j}{\sin J} = \frac{k}{\sin K}[/tex]
So, we have:
[tex]\frac{420}{\sin 43} = \frac{480}{\sin K}[/tex]
Cross multiply
[tex]\sin K * 420 = \sin 43 * 480[/tex]
Make sin K the subject
[tex]\sin K= \frac{\sin 43 * 480}{420}[/tex]
[tex]\sin K= \frac{0.6820 * 480}{420}[/tex]
[tex]\sin K= 0.7794[/tex]
Take arc sin of both sides
[tex]K = sin^{-1}(0.7794)[/tex]
[tex]K \approx 51.2^\circ[/tex]
Recall that: [tex]0^\circ \le K \le 180^\circ[/tex]
The other possible value of K is:
[tex]K \approx 180 - 51.2[/tex]
[tex]K \approx 128.8^\circ[/tex]
