Answer:
[tex]\sf AB = 2, BC = 6, AC = 7 [/tex]
Step-by-step explanation:
To determine which measures of the sides of triangle ABC angle B is the largest angle of the triangle, we can use the Law of Cosines to calculate the cosine of each angle in the triangle. The largest angle will correspond to the smallest cosine value.
The Law of Cosines states:
[tex]\large\boxed{\boxed{\sf c^2 = a^2 + b^2 - 2ab \cos C}} [/tex]
where
Let's calculate the cosines of angles A, B, and C for each set of side lengths.
For the first set of side lengths:
[tex]\sf AB = 2, BC = 6, AC = 7 [/tex]
Let [tex]\sf a = 2 [/tex], [tex]\sf b = 6 [/tex], [tex]\sf c = 7 [/tex].
Using the Law of Cosines for angle B:
[tex]\sf c^2 = a^2 + b^2 - 2ab \cos C [/tex]
[tex]\sf 7^2 = 2^2 + 6^2 - 2 \times 2 \times 6 \cos B [/tex]
[tex]\sf 49 = 4 + 36 - 24 \cos B [/tex]
[tex]\sf 49 = 40 - 24 \cos B [/tex]
[tex]\sf 24 \cos B = -9 [/tex]
[tex]\sf \cos B = \dfrac{-9}{24} = -\dfrac{3}{8} [/tex]
Since [tex]\sf \cos B [/tex] is negative, angle B is obtuse and the largest angle.
[tex]\dotfill[/tex]
For the second set of side lengths:
[tex]\sf AB = 6, BC = 12, AC = 8 [/tex]
Let [tex]\sf a = 6 [/tex], [tex]\sf b = 12 [/tex], [tex]\sf c = 8 [/tex].
Using the Law of Cosines for angle B:
[tex]\sf c^2 = a^2 + b^2 - 2ab \cos C [/tex]
[tex]\sf 8^2 = 6^2 + 12^2 - 2 \times 6 \times 12 \cos B [/tex]
[tex]\sf 64 = 36 + 144 - 144 \cos B [/tex]
[tex]\sf 64 = 180 - 144 \cos B [/tex]
[tex]\sf 144 \cos B = 116 [/tex]
[tex]\sf \cos B = \dfrac{116}{144} = \dfrac{29}{36} [/tex]
Since [tex]\sf \cos B [/tex] is positive, angle B is acute and not the largest angle.
[tex]\dotfill[/tex]
For the third set of side lengths:
[tex]\sf AB = 16, BC = 9, AC = 10 [/tex]
Let [tex]\sf a = 16 [/tex], [tex]\sf b = 9 [/tex], [tex]\sf c = 10 [/tex].
Using the Law of Cosines for angle B:
[tex]\sf c^2 = a^2 + b^2 - 2ab \cos C [/tex]
[tex]\sf 10^2 = 16^2 + 9^2 - 2 \times 16 \times 9 \cos B [/tex]
[tex]\sf 100 = 256 + 81 - 288 \cos B [/tex]
[tex]\sf 100 = 337 - 288 \cos B [/tex]
[tex]\sf 288 \cos B = 237 [/tex]
[tex]\sf \cos B = \dfrac{237}{288} [/tex]
Since [tex]\sf \cos B [/tex] is positive, angle B is acute and not the largest angle.
Therefore, among the given sets of side lengths:
For [tex]\sf AB = 2, BC = 6, AC = 7 [/tex], angle B is the largest angle.
For [tex]\sf AB = 6, BC = 12, AC = 8 [/tex] and [tex]\sf AB = 16, BC = 9, AC = 10 [/tex], angle B is not the largest angle in the triangle.
