Question

1. three sides of a triangle are 2, $sqrt{7}$, and 3. is the triangle acute, right, or obtuse? show your work. justify your answer.

Explanation:

Step1: Recall the cosine - law

Let \(a = 2\), \(b=\sqrt{7}\), \(c = 3\). The cosine - law is \(c^{2}=a^{2}+b^{2}-2ab\cos C\), \(a^{2}=b^{2}+c^{2}-2bc\cos A\), \(b^{2}=a^{2}+c^{2}-2ac\cos B\). We can use the largest - side rule. The largest side is \(c = 3\). Then, by the cosine - law, \(\cos C=\frac{a^{2}+b^{2}-c^{2}}{2ab}\).

Step2: Substitute the values

Substitute \(a = 2\), \(b=\sqrt{7}\), \(c = 3\) into the formula for \(\cos C\).
\[

$$\begin{align*} \cos C&=\frac{2^{2}+(\sqrt{7})^{2}-3^{2}}{2\times2\times\sqrt{7}}\\ &=\frac{4 + 7-9}{4\sqrt{7}}\\ &=\frac{2}{4\sqrt{7}}\\ &=\frac{1}{2\sqrt{7}}>0 \end{align*}$$

\]

Step3: Determine the type of triangle

Since \(\cos C>0\), angle \(C\) is acute. And since the largest angle of the triangle is acute, the triangle is acute.

Answer:

The triangle is acute.