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Sagot :
To determine which angle in the triangle [tex]\( \triangle ABC \)[/tex] has the greatest measure, we can employ the Law of Cosines.
Given:
- [tex]\( BC = a = 9 \)[/tex]
- [tex]\( AB = c = 7 \)[/tex]
- [tex]\( AC = b = 13 \)[/tex]
The Law of Cosines states:
[tex]\[ c^2 = a^2 + b^2 - 2ab \cos(C) \][/tex]
[tex]\[ b^2 = a^2 + c^2 - 2ac \cos(B) \][/tex]
[tex]\[ a^2 = b^2 + c^2 - 2bc \cos(A) \][/tex]
First, let's find the cosine of each angle.
### Find [tex]\( \cos(\angle C) \)[/tex]:
[tex]\[ \cos(\angle C) = \frac{a^2 + b^2 - c^2}{2ab}\][/tex]
Substitute in the given side lengths:
[tex]\[ \cos(\angle C) = \frac{9^2 + 13^2 - 7^2}{2 \cdot 9 \cdot 13} \][/tex]
[tex]\[ \cos(\angle C) = \frac{81 + 169 - 49}{234} \][/tex]
[tex]\[ \cos(\angle C) = \frac{201}{234} \][/tex]
[tex]\[ \cos(\angle C) = \frac{67}{78} \approx 0.85897 \][/tex]
### Find [tex]\( \cos(\angle B) \)[/tex]:
[tex]\[ \cos(\angle B) = \frac{a^2 + c^2 - b^2}{2ac} \][/tex]
Substitute in the given side lengths:
[tex]\[ \cos(\angle B) = \frac{9^2 + 7^2 - 13^2}{2 \cdot 9 \cdot 7} \][/tex]
[tex]\[ \cos(\angle B) = \frac{81 + 49 - 169}{126} \][/tex]
[tex]\[ \cos(\angle B) = \frac{-39}{126} \][/tex]
[tex]\[ \cos(\angle B) = -\frac{13}{42} \approx -0.30952 \][/tex]
### Find [tex]\( \cos(\angle A) \)[/tex]:
[tex]\[ \cos(\angle A) = \frac{b^2 + c^2 - a^2}{2bc} \][/tex]
Substitute in the given side lengths:
[tex]\[ \cos(\angle A) = \frac{13^2 + 7^2 - 9^2}{2 \cdot 13 \cdot 7} \][/tex]
[tex]\[ \cos(\angle A) = \frac{169 + 49 - 81}{182} \][/tex]
[tex]\[ \cos(\angle A) = \frac{137}{182} \][/tex]
[tex]\[ \cos(\angle A) = \frac{137}{182} \approx 0.75275 \][/tex]
Now, to find the angles, we need to take the inverse cosine of each value found:
### Compute the angles:
[tex]\[ \angle C = \cos^{-1}(0.85897) \approx 30.8^\circ \][/tex]
[tex]\[ \angle B = \cos^{-1}(-0.30952) \approx 108.03^\circ \][/tex]
[tex]\[ \angle A = \cos^{-1}(0.75275) \approx 41.17^\circ \][/tex]
By comparison, [tex]\( \angle B \)[/tex] has the greatest measure at approximately [tex]\( 108.03^\circ \)[/tex].
Therefore, the angle in the triangle with the greatest measure is
[tex]\[ \boxed{\angle B} \][/tex]
Given:
- [tex]\( BC = a = 9 \)[/tex]
- [tex]\( AB = c = 7 \)[/tex]
- [tex]\( AC = b = 13 \)[/tex]
The Law of Cosines states:
[tex]\[ c^2 = a^2 + b^2 - 2ab \cos(C) \][/tex]
[tex]\[ b^2 = a^2 + c^2 - 2ac \cos(B) \][/tex]
[tex]\[ a^2 = b^2 + c^2 - 2bc \cos(A) \][/tex]
First, let's find the cosine of each angle.
### Find [tex]\( \cos(\angle C) \)[/tex]:
[tex]\[ \cos(\angle C) = \frac{a^2 + b^2 - c^2}{2ab}\][/tex]
Substitute in the given side lengths:
[tex]\[ \cos(\angle C) = \frac{9^2 + 13^2 - 7^2}{2 \cdot 9 \cdot 13} \][/tex]
[tex]\[ \cos(\angle C) = \frac{81 + 169 - 49}{234} \][/tex]
[tex]\[ \cos(\angle C) = \frac{201}{234} \][/tex]
[tex]\[ \cos(\angle C) = \frac{67}{78} \approx 0.85897 \][/tex]
### Find [tex]\( \cos(\angle B) \)[/tex]:
[tex]\[ \cos(\angle B) = \frac{a^2 + c^2 - b^2}{2ac} \][/tex]
Substitute in the given side lengths:
[tex]\[ \cos(\angle B) = \frac{9^2 + 7^2 - 13^2}{2 \cdot 9 \cdot 7} \][/tex]
[tex]\[ \cos(\angle B) = \frac{81 + 49 - 169}{126} \][/tex]
[tex]\[ \cos(\angle B) = \frac{-39}{126} \][/tex]
[tex]\[ \cos(\angle B) = -\frac{13}{42} \approx -0.30952 \][/tex]
### Find [tex]\( \cos(\angle A) \)[/tex]:
[tex]\[ \cos(\angle A) = \frac{b^2 + c^2 - a^2}{2bc} \][/tex]
Substitute in the given side lengths:
[tex]\[ \cos(\angle A) = \frac{13^2 + 7^2 - 9^2}{2 \cdot 13 \cdot 7} \][/tex]
[tex]\[ \cos(\angle A) = \frac{169 + 49 - 81}{182} \][/tex]
[tex]\[ \cos(\angle A) = \frac{137}{182} \][/tex]
[tex]\[ \cos(\angle A) = \frac{137}{182} \approx 0.75275 \][/tex]
Now, to find the angles, we need to take the inverse cosine of each value found:
### Compute the angles:
[tex]\[ \angle C = \cos^{-1}(0.85897) \approx 30.8^\circ \][/tex]
[tex]\[ \angle B = \cos^{-1}(-0.30952) \approx 108.03^\circ \][/tex]
[tex]\[ \angle A = \cos^{-1}(0.75275) \approx 41.17^\circ \][/tex]
By comparison, [tex]\( \angle B \)[/tex] has the greatest measure at approximately [tex]\( 108.03^\circ \)[/tex].
Therefore, the angle in the triangle with the greatest measure is
[tex]\[ \boxed{\angle B} \][/tex]
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