Question

If the angles of triangle are in the ratio \[4:1:1\] then the ratio of the longest side to the perimeter is
A) \[\sqrt 3 :\left( {2 + \sqrt 3 } \right)\]
B) \[1:6\]
C) \[1:\left( {2 + \sqrt 3 } \right)\]
D) \[2:3\]

Answer 1

Hint: Find the common factor in the ratios of angles which will lead you to find the angles, after that use the formula \[{a^2} = {b^2} + c{}^2 - 2bc\cos A\] to find the value of a which is the length of the opposite side of the angle A.
Complete step-by-step answer:
Given, ratio of angles are \[4:1:1\]
Now let us take that the common factor in the ratios is y and we know that the sum of all angles in a triangle is equal to \[{180^ \circ }\]
Therefore the angles become 4y, y, y let us add them all and equate it with \[{180^ \circ }\] to find the value of y.
\[\begin{array}{l}
 \Rightarrow 4y + y + y = {180^ \circ }\\
 \Rightarrow 6y = {180^ \circ }\\
 \Rightarrow y = \dfrac{{{{180}^ \circ }}}{6}\\
 \Rightarrow y = {30^ \circ }
\end{array}\]
\[\begin{array}{l}
\therefore \angle A = 4y = {120^ \circ }\\
\angle B = y = {30^ \circ }\\
\angle C = y = {30^ \circ }
\end{array}\]
Let angles A, B, C be the angles opposite to the side a, b, c respectively.
Thus the ratio of the longest side to the perimeter of the triangle will become \[\dfrac{a}{{a + b + c}}\]
Let \[b = c = x\]
\[\begin{array}{l}
 \Rightarrow {a^2} = {b^2} + {c^2} - 2bc\cos A\\
 \Rightarrow {a^2} = {x^2} + {x^2} - 2{x^2}\cos A\\
 \Rightarrow {a^2} = 2{x^2}(1 - \cos A)\\
 \Rightarrow {a^2} = 4{x^2}{\sin ^2}\dfrac{A}{2}\\
 \Rightarrow a = 2x\sin \dfrac{A}{2}\\
 \Rightarrow a = 2x\sin {60^ \circ } = \sqrt 3 x
\end{array}\]
Thus the ratio is
\[\dfrac{a}{{a + b + c}} = \dfrac{{\sqrt 3 x}}{{x + x + \sqrt 3 x}} = \dfrac{{\sqrt 3 }}{{2 + \sqrt 3 }}\]
Therefore option A is the correct option.

Note: It must be noted that \[1 - \cos A = {\sin ^2}\dfrac{A}{2} + {\cos ^2}\dfrac{A}{2} - {\cos ^2}\dfrac{A}{2} + {\sin ^2}\dfrac{A}{2} = 2{\sin ^2}\dfrac{A}{2}\] I have directly used the result while doing the solution.