[tex]\begin{gathered} X=50\text{ \degree} \\ \text{Area}_{triangle}=67.41\text{ square units} \end{gathered}[/tex]
Explanation
the law of cosines states that
[tex]c^2=a^2-b^2-2abCosC[/tex]
then
Step 1
let
c=16
C=X
a=22
b=8
Now, replace
[tex]\begin{gathered} c^2=a^2-b^2-2abCosC \\ 16^2=22^2-8^2-2\cdot16\cdot8\cdot\text{CosX} \\ 256=484-64-256CosX \\ 256=420-256\text{CosX} \end{gathered}[/tex]
Step 2
now, isolate x
[tex]\begin{gathered} 256=420-256\text{CosX} \\ subtract\text{ 420 in both sides} \\ 256-420=420-256\text{CosX-420} \\ -164=-256\text{CosX} \\ \text{divide both sides by -256} \\ \frac{-164}{-256}=\frac{-256\text{CosX}}{-256} \\ \frac{41}{64}=\text{Cos X} \\ \text{Get the inverse cosine in both sides} \\ \cos ^{-1}(\frac{41}{64})=\cos ^{-1}(\cos X) \\ 50.162=X \\ \text{rounded to the whole degre}e \\ X=50\text{ \degree} \end{gathered}[/tex]
Step 3
Now, find the area of the triangle, using:
[tex]\text{Area}_{triangle}=\frac{1}{2}ab\cdot\sin \text{ C}[/tex]
then, replace
[tex]\begin{gathered} \text{Area}_{triangle}=\frac{1}{2}ab\cdot\sin \text{ C} \\ \text{Area}_{triangle}=\frac{1}{2}(22\cdot8)\cdot\sin 50 \\ \text{Area}_{triangle}=88\cdot\sin 50 \\ \text{Area}_{triangle}=88\cdot0.76 \\ \text{Area}_{triangle}=67.41\text{ square units} \end{gathered}[/tex]
I hope this helps you
