The value of the angle Q in the given diagram is 30.75 degrees.
The previous problem, this one requires the application of the Law of Cosines.
We have to determine the angle Q when you know the lengths of all 3 sides of the triangle.
What is the law of Cosines?
a^2 = b^2 + c^2 - 2bc cos A
[tex]32^2+64^2-2\cdot \:32\cdot \:64\cos \left(Q\right)=40^2[/tex]
[tex]32^2+64^2-4096\cos \left(Q\right)=40^2[/tex]
[tex]1024+64^2-4096\cos \left(Q\right)=40^2[/tex]
[tex]1024+4096-4096\cos \left(Q\right)=40^2[/tex]
[tex]-4096\cos \left(Q\right)+5120=40^2[/tex]
[tex]-4096\cos \left(Q\right)+5120=1600[/tex]
[tex]-4096\cos \left(Q\right)=-3520[/tex]
[tex]\frac{-4096\cos \left(Q\right)}{-4096}=\frac{-3520}{-4096}[/tex]
[tex]\cos \left(Q\right)=\frac{55}{64}[/tex]
[tex]Q=30.75^0[/tex]
Solve for cos Q, and then find Q in degrees and Q in radians.
Therefore we get the value of the angle Q is 30.75 degrees.
To learn more about the angle visit:
https://brainly.com/question/25770607
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