Question

question 49 of 50
if a triangle meets the criteria of c² < a² + b², then the triangle is acute.
∘ false
∘ true

Explanation:

Step1: Recall the Law of Cosines

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

Step2: Analyze the given inequality \(c^{2}

Substitute the Law of Cosines into the inequality: \(a^{2}+b^{2}-2ab\cos C < a^{2}+b^{2}\).
Subtract \(a^{2}+b^{2}\) from both sides: \(- 2ab\cos C<0\).
Since \(a\) and \(b\) are side lengths of a triangle, \(a>0\) and \(b > 0\), so \(ab>0\). Then we can divide both sides by \(- 2ab\) (and reverse the inequality sign) to get \(\cos C>0\).

Step3: Determine the type of angle \(C\)

The cosine of an angle is positive when the angle is between \(0^{\circ}\) and \(90^{\circ}\) (acute angle). In a triangle, if the largest angle is acute, then the triangle is acute. By the Law of Cosines, if \(c\) is the longest side, and \(c^{2}

Answer:

True (the option with "True" text)