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Question
Answer 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.
Solution
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
\[\cos B = \frac{c^2 + a^2 - b^2}{2ca}\]
\[ \Rightarrow \cos B = \frac{\left( \sqrt{3} - 1 \right)^2 + 2^2 - \left( \sqrt{6} \right)^2}{2 \times \left( \sqrt{3} - 1 \right) \times 2}\]
\[ \Rightarrow \cos B = \frac{3 + 1 - 2\sqrt{3} + 4 - 6}{4\left( \sqrt{3} - 1 \right)}\]
\[\Rightarrow \cos B = \frac{2 - 2\sqrt{3}}{4\left( \sqrt{3} - 1 \right)} = \frac{- 2\left( \sqrt{3} - 1 \right)}{4\left( \sqrt{3} - 1 \right)}\]
\[ \Rightarrow \cos B = - \frac{1}{2} = \cos120°\]
\[ \Rightarrow B = 120°\]
Hence, the measure of its greatest angle is 120º.
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