Question

If the sides of a triangle are in the ratio \[2:\sqrt 6 :\left( {\sqrt 3 + 1} \right)\], then find the largest angle of the triangle.
A. \[{60^ \circ }\]
B. \[{75^ \circ }\]
C. \[{90^ \circ }\]
D. \[{120^ \circ }\]

Answer 1

Hint: From the ratio, we will find the length of the sides. Using the sides and cosine law, we will find all angles of the triangle. Then we will find the largest angle.

Formula used:
Cosine Law:
\[{a^2} = {b^2} + {c^2} - 2bc\cos A\]
\[{b^2} = {a^2} + {c^2} - 2ac\cos B\]
\[{c^2} = {a^2} + {b^2} - 2ab\cos C\]

Complete step by step solution:
Given that, the ratio of sides of a triangle is \[2:\sqrt 6 :\left( {\sqrt 3 + 1} \right)\].
Assume that the sides of the triangle are \[a = 2k\], \[b = \sqrt 6 k\], and \[c = \left( {\sqrt 3 + 1} \right)k\]
Now calculating the angle A using cosine law \[{a^2} = {b^2} + {c^2} - 2bc\cos A\]:
\[{\left( {2k} \right)^2} = {\left( {\sqrt 6 k} \right)^2} + {\left( {\sqrt 3 + 1} \right)^2}{k^2} - 2 \cdot \left( {\sqrt 6 } \right)k \cdot \left( {\sqrt 3 + 1} \right)k \cdot \cos A\]
\[ \Rightarrow 4{k^2} = 6{k^2} + \left( {3 + 2\sqrt 3 + 1} \right){k^2} - 2 \cdot \sqrt 6 \left( {\sqrt 3 + 1} \right){k^2}\cos A\]
Cancel out \[{k^2}\] from both sides
\[ \Rightarrow 4 = 6 + 3 + 2\sqrt 3 + 1 - 2\sqrt 6 \left( {\sqrt 3 + 1} \right)\cos A\]
\[ \Rightarrow 4 = 6 + 3 + 2\sqrt 3 + 1 - 2\sqrt 6 \left( {\sqrt 3 + 1} \right)\cos A\]
Simplify the above equation:
\[ \Rightarrow 6 + 3 + 2\sqrt 3 + 1 - 4 = 2\sqrt 6 \left( {\sqrt 3 + 1} \right)\cos A\]
\[ \Rightarrow 6 + 2\sqrt 3 = 2\sqrt 6 \left( {\sqrt 3 + 1} \right)\cos A\]
Divide both sides by \[2\sqrt 6 \left( {\sqrt 3 + 1} \right)\]
\[ \Rightarrow \dfrac{{2\sqrt 3 \left( {\sqrt 3 + 1} \right)}}{{2\sqrt 6 \left( {\sqrt 3 + 1} \right)}} = \cos A\]
Cancel out common terms
\[ \Rightarrow \dfrac{{\sqrt 3 }}{{\sqrt 3 \cdot \sqrt 2 }} = \cos A\]
\[ \Rightarrow \dfrac{1}{{\sqrt 2 }} = \cos A\]
\[ \Rightarrow A = {45^ \circ }\]
Now calculating the angle B using cosine law \[{b^2} = {a^2} + {c^2} - 2ac\cos B\]:
\[{\left( {\sqrt 6 k} \right)^2} = {\left( {2k} \right)^2} + {\left( {\sqrt 3 + 1} \right)^2}{k^2} - 2 \cdot \left( {2k} \right) \cdot \left( {\sqrt 3 + 1} \right)k \cdot \cos B\]
\[ \Rightarrow 6{k^2} = 4{k^2} + \left( {3 + 2\sqrt 3 + 1} \right){k^2} - 4\left( {\sqrt 3 + 1} \right){k^2}\cos B\]
Cancel out \[{k^2}\] from both sides
\[ \Rightarrow 6 = 4 + \left( {3 + 2\sqrt 3 + 1} \right) - 4\left( {\sqrt 3 + 1} \right)\cos B\]
\[ \Rightarrow 4 + 3 + 2\sqrt 3 + 1 - 6 = 4\left( {\sqrt 3 + 1} \right)\cos B\]
Simplify the above equation:
\[ \Rightarrow 2 + 2\sqrt 3 = 4\left( {\sqrt 3 + 1} \right)\cos B\]
\[ \Rightarrow 2\left( {1 + \sqrt 3 } \right) = 4\left( {\sqrt 3 + 1} \right)\cos B\]
Divide both sides by \[4\left( {\sqrt 3 + 1} \right)\]
\[ \Rightarrow \dfrac{{2\left( {1 + \sqrt 3 } \right)}}{{4\left( {\sqrt 3 + 1} \right)}} = \cos B\]
Cancel out common terms
\[ \Rightarrow \dfrac{1}{2} = \cos B\]
\[ \Rightarrow B = {60^ \circ }\]
Now calculating the angle C using cosine law \[{c^2} = {a^2} + {b^2} - 2ab\cos C\]:
\[{\left( {\sqrt 3 + 1} \right)^2}{k^2} = {\left( {2k} \right)^2} + {\left( {\sqrt 6 k} \right)^2} - 2 \cdot \left( {2k} \right) \cdot \left( {\sqrt 6 k} \right) \cdot \cos C\]
\[ \Rightarrow \left( {3 + 2\sqrt 3 + 1} \right){k^2} = 4{k^2} + 6{k^2} - 4\sqrt 6 {k^2}\cos C\]
Cancel out \[{k^2}\] from both sides
\[ \Rightarrow \left( {3 + 2\sqrt 3 + 1} \right) = 4 + 6 - 4\sqrt 6 \cos C\]
\[ \Rightarrow \left( {4 + 2\sqrt 3 } \right) = 10 - 4\sqrt 6 \cos C\]
Simplify the above equation:
\[ \Rightarrow 10 - 4 - 2\sqrt 3 = 4\sqrt 6 \cos C\]
\[ \Rightarrow 6 - 2\sqrt 3 = 4\sqrt 6 \cos C\]
Divide both sides by \[4\sqrt 6 \]
\[ \Rightarrow \dfrac{{2\sqrt 3 \left( {\sqrt 3 - 1} \right)}}{{4\sqrt 6 }} = \cos C\]
Cancel out common terms
\[ \Rightarrow \dfrac{{\left( {\sqrt 3 - 1} \right)}}{{2\sqrt 2 }} = \cos C\]
\[ \Rightarrow C = {75^ \circ }\] [ since \[\dfrac{{\left( {\sqrt 3 - 1} \right)}}{{2\sqrt 2 }} = \cos {75^ \circ }\]]
Hence the largest angle of the triangle is \[{75^ \circ }\].
Hence option B is the correct option.

Note: Students often make mistakes to solve this question. They calculate only one angle and take that angle as the largest angle. But here we need to find all angles, then identify which one is the largest angle.