Given a triangle with sides [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex], where [tex]\(\theta\)[/tex] is the angle opposite the side of length [tex]\(a\)[/tex], and knowing that [tex]\(\cos \theta > 0\)[/tex], we need to determine what must be true among the given options:
A. [tex]\(a^2 + b^2 > c^2\)[/tex]
B. [tex]\(b^2 + c^2 > a^2\)[/tex]
C. [tex]\(b^2 + c^2 < a^2\)[/tex]
D. [tex]\(a^2 + b^2 = c^2\)[/tex]
### Step-by-Step Solution:
1. Understanding [tex]\(\cos \theta > 0\)[/tex]:
[tex]\(\cos \theta > 0\)[/tex] implies that the angle [tex]\(\theta\)[/tex] is such that [tex]\(\theta\)[/tex] lies in a range where the cosine function is positive. This happens in two conditions:
- [tex]\(0^\circ < \theta < 90^\circ\)[/tex] (1st Quadrant)
- [tex]\(270^\circ < \theta < 360^\circ\)[/tex] (4th Quadrant)
In the context of a triangle, angles can only be between [tex]\(0^\circ\)[/tex] and [tex]\(180^\circ\)[/tex]. Hence, [tex]\(\theta\)[/tex] must be an acute angle ([tex]\(0^\circ < \theta < 90^\circ\)[/tex]).
2. Law of Cosines Application:
For any triangle with sides [tex]\(a\)[/tex], [tex]\(b\)[/tex], [tex]\(c\)[/tex] and angle [tex]\(\theta\)[/tex] opposite to side [tex]\(a\)[/tex]:
[tex]\[
\cos \theta = \frac{b^2 + c^2 - a^2}{2bc}
\][/tex]
Knowing [tex]\(\cos \theta > 0\)[/tex], we can infer that:
[tex]\[
\frac{b^2 + c^2 - a^2}{2bc} > 0
\][/tex]
This inequality simplifies (since [tex]\(2bc > 0\)[/tex] for positive side lengths):
[tex]\[
b^2 + c^2 - a^2 > 0 \implies b^2 + c^2 > a^2
\][/tex]
3. Conclusion:
This implies that the inequality [tex]\(b^2 + c^2 > a^2\)[/tex] must hold true when [tex]\(\cos \theta > 0\)[/tex] in any triangle.
Therefore, the correct option is:
[tex]\[
\boxed{\text{B. } b^2 + c^2 > a^2}
\][/tex]
