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To find the measure of [tex]\(\angle Q\)[/tex], the smallest angle in a triangle with side lengths [tex]\(a = 4\)[/tex], [tex]\(b = 5\)[/tex], and [tex]\(c = 6\)[/tex], we can use the Law of Cosines. The Law of Cosines relates one angle of a triangle to the lengths of its sides and can be written as:
[tex]\[ a^2 = b^2 + c^2 - 2bc \cos(A) \][/tex]
We can rearrange this formula to solve for [tex]\(\cos(A)\)[/tex]:
[tex]\[ \cos(A) = \frac{b^2 + c^2 - a^2}{2bc} \][/tex]
We will also use the same formula to find the cosine of the other two angles. Given the side lengths, let's compute each angle step by step.
1. Finding [tex]\(\angle A\)[/tex] where [tex]\(a = 4\)[/tex], [tex]\(b = 5\)[/tex], and [tex]\(c = 6\)[/tex]:
[tex]\[ \cos(A) = \frac{5^2 + 6^2 - 4^2}{2 \cdot 5 \cdot 6} = \frac{25 + 36 - 16}{60} = \frac{45}{60} = 0.75 \][/tex]
[tex]\[ \angle A = \arccos(0.75) \][/tex]
Using a calculator, we find:
[tex]\[ \angle A \approx 41.41^{\circ} \][/tex]
2. Finding [tex]\(\angle B\)[/tex] where [tex]\(a = 4\)[/tex], [tex]\(b = 5\)[/tex], and [tex]\(c = 6\)[/tex]:
[tex]\[ \cos(B) = \frac{4^2 + 6^2 - 5^2}{2 \cdot 4 \cdot 6} = \frac{16 + 36 - 25}{48} = \frac{27}{48} = 0.5625 \][/tex]
[tex]\[ \angle B = \arccos(0.5625) \][/tex]
Using a calculator, we find:
[tex]\[ \angle B \approx 55.77^{\circ} \][/tex]
3. Finding [tex]\(\angle C\)[/tex] where [tex]\(a = 4\)[/tex], [tex]\(b = 5\)[/tex], and [tex]\(c = 6\)[/tex]:
[tex]\[ \cos(C) = \frac{4^2 + 5^2 - 6^2}{2 \cdot 4 \cdot 5} = \frac{16 + 25 - 36}{40} = \frac{5}{40} = 0.125 \][/tex]
[tex]\[ \angle C = \arccos(0.125) \][/tex]
Using a calculator, we find:
[tex]\[ \angle C \approx 82.82^{\circ} \][/tex]
We now have the measures of the three angles:
[tex]\[ \angle A \approx 41.41^{\circ}, \quad \angle B \approx 55.77^{\circ}, \quad \angle C \approx 82.82^{\circ} \][/tex]
The smallest angle is [tex]\(\angle A \approx 41.41^{\circ}\)[/tex], which, when rounded to the nearest whole number, is [tex]\(41^{\circ}\)[/tex].
Therefore, the measure of [tex]\(\angle Q\)[/tex], the smallest angle, is:
[tex]\[ \boxed{41^{\circ}} \][/tex]
[tex]\[ a^2 = b^2 + c^2 - 2bc \cos(A) \][/tex]
We can rearrange this formula to solve for [tex]\(\cos(A)\)[/tex]:
[tex]\[ \cos(A) = \frac{b^2 + c^2 - a^2}{2bc} \][/tex]
We will also use the same formula to find the cosine of the other two angles. Given the side lengths, let's compute each angle step by step.
1. Finding [tex]\(\angle A\)[/tex] where [tex]\(a = 4\)[/tex], [tex]\(b = 5\)[/tex], and [tex]\(c = 6\)[/tex]:
[tex]\[ \cos(A) = \frac{5^2 + 6^2 - 4^2}{2 \cdot 5 \cdot 6} = \frac{25 + 36 - 16}{60} = \frac{45}{60} = 0.75 \][/tex]
[tex]\[ \angle A = \arccos(0.75) \][/tex]
Using a calculator, we find:
[tex]\[ \angle A \approx 41.41^{\circ} \][/tex]
2. Finding [tex]\(\angle B\)[/tex] where [tex]\(a = 4\)[/tex], [tex]\(b = 5\)[/tex], and [tex]\(c = 6\)[/tex]:
[tex]\[ \cos(B) = \frac{4^2 + 6^2 - 5^2}{2 \cdot 4 \cdot 6} = \frac{16 + 36 - 25}{48} = \frac{27}{48} = 0.5625 \][/tex]
[tex]\[ \angle B = \arccos(0.5625) \][/tex]
Using a calculator, we find:
[tex]\[ \angle B \approx 55.77^{\circ} \][/tex]
3. Finding [tex]\(\angle C\)[/tex] where [tex]\(a = 4\)[/tex], [tex]\(b = 5\)[/tex], and [tex]\(c = 6\)[/tex]:
[tex]\[ \cos(C) = \frac{4^2 + 5^2 - 6^2}{2 \cdot 4 \cdot 5} = \frac{16 + 25 - 36}{40} = \frac{5}{40} = 0.125 \][/tex]
[tex]\[ \angle C = \arccos(0.125) \][/tex]
Using a calculator, we find:
[tex]\[ \angle C \approx 82.82^{\circ} \][/tex]
We now have the measures of the three angles:
[tex]\[ \angle A \approx 41.41^{\circ}, \quad \angle B \approx 55.77^{\circ}, \quad \angle C \approx 82.82^{\circ} \][/tex]
The smallest angle is [tex]\(\angle A \approx 41.41^{\circ}\)[/tex], which, when rounded to the nearest whole number, is [tex]\(41^{\circ}\)[/tex].
Therefore, the measure of [tex]\(\angle Q\)[/tex], the smallest angle, is:
[tex]\[ \boxed{41^{\circ}} \][/tex]
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