Question

question 5 of 10
suppose a triangle has sides a, b, and c with side c the longest side, and that
$a^{2}+b^{2}>c^{2}$. let $\theta$ be the measure of the angle opposite the side of length c.
which of the following must be true? check all that apply.
a. the triangle is not a right triangle.
b. $\cos\theta<0$
c. the triangle in question is a right triangle.
d. $\theta$ is an acute angle.

Explanation:

Step1: Recall the Law of Cosines

The Law of Cosines states that for a triangle with sides \(a\), \(b\), \(c\) and the angle \(\theta\) opposite side \(c\), \(c^{2}=a^{2}+b^{2}-2ab\cos\theta\).

Step2: Analyze the given condition \(a^{2}+b^{2}>c^{2}\)

Substitute the Law of Cosines into the inequality:
\(a^{2}+b^{2}>a^{2}+b^{2}-2ab\cos\theta\)
Subtract \(a^{2}+b^{2}\) from both sides:
\(0 > - 2ab\cos\theta\)
Multiply both sides by \(- 1\) (and reverse the inequality sign):
\(0<2ab\cos\theta\)
Since \(a\) and \(b\) are side lengths of a triangle, \(a>0\) and \(b > 0\), so \(2ab>0\). Then, for \(0 < 2ab\cos\theta\) to hold, we must have \(\cos\theta>0\).

Step3: Analyze the angle \(\theta\)

If \(\cos\theta>0\), and \(\theta\) is an angle in a triangle (so \(0^{\circ}<\theta<180^{\circ}\)), then \(\theta\) must be an acute angle (because \(\cos\theta>0\) when \(0^{\circ}<\theta < 90^{\circ}\)).

Step4: Analyze if the triangle is a right triangle

For a right triangle with hypotenuse \(c\), by the Pythagorean theorem, \(a^{2}+b^{2}=c^{2}\). But we are given \(a^{2}+b^{2}>c^{2}\), so the triangle is not a right triangle.

Step5: Analyze option B

We found that \(\cos\theta>0\), so \(\cos\theta < 0\) is false.

Answer:

A. The triangle is not a right triangle.
D. \(\theta\) is an acute angle.