Answer :
To determine which angle in a triangle has the greatest measure given the side lengths [tex]\( BC = 9 \)[/tex], [tex]\( AB = 7 \)[/tex], and [tex]\( AC = 13 \)[/tex], we can use the Law of Cosines. The Law of Cosines states that for any triangle with sides [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex], and opposite angles [tex]\( A \)[/tex], [tex]\( B \)[/tex], and [tex]\( C \)[/tex] respectively:
[tex]\[ \cos(A) = \frac{b^2 + c^2 - a^2}{2bc} \][/tex]
[tex]\[ \cos(B) = \frac{a^2 + c^2 - b^2}{2ac} \][/tex]
[tex]\[ \cos(C) = \frac{a^2 + b^2 - c^2}{2ab} \][/tex]
Given the side lengths:
- [tex]\( a = AC = 13 \)[/tex]
- [tex]\( b = AB = 7 \)[/tex]
- [tex]\( c = BC = 9 \)[/tex]
We can calculate the cosine of each angle:
1. To find [tex]\(\cos(A)\)[/tex]:
[tex]\[ \cos(A) = \frac{b^2 + c^2 - a^2}{2bc} = \frac{7^2 + 9^2 - 13^2}{2 \times 7 \times 9} \][/tex]
[tex]\[ \cos(A) = \frac{49 + 81 - 169}{126} = \frac{-39}{126} = -0.3095 \][/tex]
2. To find [tex]\(\cos(B)\)[/tex]:
[tex]\[ \cos(B) = \frac{a^2 + c^2 - b^2}{2ac} = \frac{13^2 + 9^2 - 7^2}{2 \times 13 \times 9} \][/tex]
[tex]\[ \cos(B) = \frac{169 + 81 - 49}{234} = \frac{201}{234} = 0.858 \][/tex]
3. To find [tex]\(\cos(C)\)[/tex]:
[tex]\[ \cos(C) = \frac{a^2 + b^2 - c^2}{2ab} = \frac{13^2 + 7^2 - 9^2}{2 \times 13 \times 7} \][/tex]
[tex]\[ \cos(C) = \frac{169 + 49 - 81}{182} = \frac{137}{182} = 0.753 \][/tex]
Next, we convert these cosine values to angles in degrees:
- [tex]\(\angle A \approx 108.03^\circ\)[/tex]
- [tex]\(\angle B \approx 41.17^\circ\)[/tex]
- [tex]\(\angle C \approx 30.80^\circ\)[/tex]
Now we compare the angles:
- [tex]\(\angle A = 108.03^\circ\)[/tex]
- [tex]\(\angle B = 41.17^\circ\)[/tex]
- [tex]\(\angle C = 30.80^\circ\)[/tex]
The greatest angle is [tex]\(\angle A = 108.03^\circ\)[/tex].
Thus, the angle in the triangle with the greatest measure is:
[tex]\[ \boxed{\angle A} \][/tex]
[tex]\[ \cos(A) = \frac{b^2 + c^2 - a^2}{2bc} \][/tex]
[tex]\[ \cos(B) = \frac{a^2 + c^2 - b^2}{2ac} \][/tex]
[tex]\[ \cos(C) = \frac{a^2 + b^2 - c^2}{2ab} \][/tex]
Given the side lengths:
- [tex]\( a = AC = 13 \)[/tex]
- [tex]\( b = AB = 7 \)[/tex]
- [tex]\( c = BC = 9 \)[/tex]
We can calculate the cosine of each angle:
1. To find [tex]\(\cos(A)\)[/tex]:
[tex]\[ \cos(A) = \frac{b^2 + c^2 - a^2}{2bc} = \frac{7^2 + 9^2 - 13^2}{2 \times 7 \times 9} \][/tex]
[tex]\[ \cos(A) = \frac{49 + 81 - 169}{126} = \frac{-39}{126} = -0.3095 \][/tex]
2. To find [tex]\(\cos(B)\)[/tex]:
[tex]\[ \cos(B) = \frac{a^2 + c^2 - b^2}{2ac} = \frac{13^2 + 9^2 - 7^2}{2 \times 13 \times 9} \][/tex]
[tex]\[ \cos(B) = \frac{169 + 81 - 49}{234} = \frac{201}{234} = 0.858 \][/tex]
3. To find [tex]\(\cos(C)\)[/tex]:
[tex]\[ \cos(C) = \frac{a^2 + b^2 - c^2}{2ab} = \frac{13^2 + 7^2 - 9^2}{2 \times 13 \times 7} \][/tex]
[tex]\[ \cos(C) = \frac{169 + 49 - 81}{182} = \frac{137}{182} = 0.753 \][/tex]
Next, we convert these cosine values to angles in degrees:
- [tex]\(\angle A \approx 108.03^\circ\)[/tex]
- [tex]\(\angle B \approx 41.17^\circ\)[/tex]
- [tex]\(\angle C \approx 30.80^\circ\)[/tex]
Now we compare the angles:
- [tex]\(\angle A = 108.03^\circ\)[/tex]
- [tex]\(\angle B = 41.17^\circ\)[/tex]
- [tex]\(\angle C = 30.80^\circ\)[/tex]
The greatest angle is [tex]\(\angle A = 108.03^\circ\)[/tex].
Thus, the angle in the triangle with the greatest measure is:
[tex]\[ \boxed{\angle A} \][/tex]
