Sure, let's find the measure of [tex]\(\angle Q\)[/tex] in a triangle with side lengths 4, 5, and 6 using the Law of Cosines.
The Law of Cosines states:
[tex]\[ a^2 = b^2 + c^2 - 2bc \cos(A) \][/tex]
Given the sides of the triangle:
[tex]\[ a = 4 \][/tex]
[tex]\[ b = 5 \][/tex]
[tex]\[ c = 6 \][/tex]
We need to find [tex]\(\angle A\)[/tex], which in this context is [tex]\(\angle Q\)[/tex].
First, rewrite the Law of Cosines formula to solve for [tex]\(\cos(A)\)[/tex]:
[tex]\[ \cos(A) = \frac{b^2 + c^2 - a^2}{2bc} \][/tex]
Substitute the given side lengths into the equation:
[tex]\[ \cos(\angle Q) = \frac{5^2 + 6^2 - 4^2}{2 \cdot 5 \cdot 6} \][/tex]
[tex]\[ \cos(\angle Q) = \frac{25 + 36 - 16}{60} \][/tex]
[tex]\[ \cos(\angle Q) = \frac{45}{60} \][/tex]
[tex]\[ \cos(\angle Q) = \frac{3}{4} \][/tex]
Now, find the angle whose cosine is [tex]\(\frac{3}{4}\)[/tex]:
[tex]\[ \angle Q = \cos^{-1}\left(\frac{3}{4}\right) \][/tex]
Converting this value to degrees (you would usually use a calculator for this step):
[tex]\[ \angle Q \approx 41^\circ \][/tex]
Hence, the measure of [tex]\(\angle Q\)[/tex], rounded to the nearest whole degree, is:
[tex]\[ 41^\circ \][/tex]
Therefore, the measure of [tex]\(\angle Q\)[/tex] in the triangle is [tex]\(41^\circ\)[/tex]. Thus, the appropriate answer choice is:
[tex]\[ \boxed{41^\circ} \][/tex]
