The correct option is: 53.9°
Explanation
According to the below diagram, [tex]\triangle UVW[/tex] is acute triangle, in which [tex]\overline{UV}= 7.6 inches, \overline{VW}= 8.6 inches[/tex] and [tex]\overline{WU}=7.4 inches[/tex]
If we apply cosine rule here, then we will get.....
[tex]cos(V) = \frac{(UV)^2+(VW)^2 -(WU)^2}{2(UV)(VW)}[/tex]
Now, plugging the given values, we will get....
[tex]cos(V)=\frac{(7.6)^2+(8.6)^2-(7.4)^2}{2(7.6)(8.6)}\\ \\ cos(V)= \frac{76.96}{130.72}=0.58873...\\ \\ V= cos^-^1(0.58873...)=53.932.... \approx 53.9 degree[/tex]
So, the measure of [tex]\angle V[/tex] is 53.9°
