Find the value of x Round to the nearest tenth
Pls help :(

Step-by-step explanation:
Hey, there!!
Let's simply work with it,
Here,
In figure ABC,
let a= 14
b= 19
and c= 8.
Using cosine rule,
[tex] {c}^{2} = {a}^{2} + {b}^{2} - 2ab.cosc[/tex]
putting value of all we get,
[tex] {8}^{2} = {14}^{2} + {19}^{2} - 2 \times 14 \times 19 \times cosx[/tex]
Now, simplifying them we get,
[tex]64 = 196 + 361 - 532.cosx[/tex]
[tex]532cosx = 557 - 64[/tex]
[tex]or \: 532cosx = 493[/tex]
[tex]or \: cosx = \frac{493}{532} [/tex]
cosx= 0.926691
Putting cos inverse,
[tex]x = {cos}^{ - 1} (0.926691) [/tex]
Therefore, x= 22.0752°
By rounding off we get,
x= 22° { as 0 is smaller than 5 itsvalue dont change}.
Hope it helps....