Question

Answer each of the following questions in one word or one sentence or as per exact requirement of the question.

If the sides of a triangle are proportional to 2, $\sqrt{6}$ and $\sqrt{3}-1$, find the measure of its greatest angle.

Answer 1

Let ∆ABC be the triangle such that a = 2, b = $\sqrt{6}$ and c = $\sqrt{3}-1$.

Clearly, b > a > c. Then,

$\angle$B is the greatest angle of ∆ABC. (Greatest side has greatest angle opposite to it)

Using cosine formula, we have

$$\begin{gathered} \cos B=\frac{c^2+a^2-b^2}{2ca} \\ \Rightarrow \cos B=\frac{{\left(\sqrt{3}-1\right)}^2+2^2-{\left(\sqrt{6}\right)}^2}{2\times \left(\sqrt{3}-1\right)\times 2} \\ \Rightarrow \cos B=\frac{3+1-2\sqrt{3}+4-6}{4\left(\sqrt{3}-1\right)} \end{gathered}$$
$$\begin{gathered} \Rightarrow \cos B=\frac{2-2\sqrt{3}}{4\left(\sqrt{3}-1\right)}=\frac{-2\left(\sqrt{3}-1\right)}{4\left(\sqrt{3}-1\right)} \\ \Rightarrow \cos B=-\frac{1}{2}=\cos 120° \\ \Rightarrow B=120° \end{gathered}$$

Hence, the measure of its greatest angle is 120º.