Question

If the angles of a triangle are in the ratio 1 : 1 : 4, then the ratio of the perimeter of the triangle to its largest side is
A. $ \sqrt{3}+2:\sqrt{3} $
B. 3 : 2
C. $ \sqrt{3}-2:\sqrt{2} $
D. $ \sqrt{3}+2:\sqrt{2} $

Answer 1

Hint:We will use the property of a triangle that the sum of the interior angles of a triangle is equal to $ 180{}^\circ $ to find each angle. Then we will use the sine rule for triangle according to which, if we have a triangle ABC with a, b and c as the corresponding sides of the triangle, then, $ \dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C} $ . We will also use the property that the side opposite to the largest angle is the largest side of the triangle.

Complete step-by-step answer:
We have been given that the angles of a triangle are in the ratio of 1 : 1 : 4. To solve this question, let us start by assuming the angles to be, K, K and 4K. We know that the sum of the angles of a triangle is equal to $ 180{}^\circ $ . So, we can write as,
 $ \begin{align}
  & K+K+4K=180{}^\circ \\
 & \Rightarrow 6K=180{}^\circ \\
 & \Rightarrow K=\dfrac{180{}^\circ }{6} \\
 & \Rightarrow K=30{}^\circ \\
\end{align} $
Therefore, the angles of the triangle are $ 30{}^\circ ,30{}^\circ ,120{}^\circ $ . Let the triangle be ABC, where $ \angle A=120{}^\circ ,\angle B=\angle C=30{}^\circ $ and the side corresponding to the angles be, a, b, and c.

We will use the sine rule for a triangle which states that, $ \dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C} $ . So, applying this rule, we will get,
 $ \dfrac{a}{\sin 120{}^\circ }=\dfrac{b}{\sin 30{}^\circ }=\dfrac{c}{\sin 30{}^\circ } $
Let their ratios be,
 $ \dfrac{a}{\sin 120{}^\circ }=\dfrac{b}{\sin 30{}^\circ }=\dfrac{c}{\sin 30{}^\circ }=K $
Now, let us take each one separately. So, we get,
 $ a=K\sin 120{}^\circ $
We know that the value of $ \sin 120{}^\circ =\dfrac{\sqrt{3}}{2} $ , so, $ a=\dfrac{\sqrt{3}}{2}K $ .
Now, for, $ a=K\sin 30{}^\circ $
We know that value of $ \sin 30{}^\circ =\dfrac{1}{2} $ , so $ b=\dfrac{1}{2}K $ .
Since A and C are the same, $ c=\dfrac{1}{2}K $ also. We know that the perimeter of a triangle is the sum of their sides, or perimeter (S) = a + b + c. We have got the sides of the triangle as, $ a=\dfrac{\sqrt{3}}{2}K $ , $ b=\dfrac{1}{2}K $ and $ c=\dfrac{1}{2}K $ . So, the perimeter of the triangle will be,
 $ \begin{align}
  & S=\dfrac{\sqrt{3}}{2}K+\dfrac{K}{2}+\dfrac{K}{2} \\
 & \Rightarrow S=\dfrac{\sqrt{3}K+K+K}{2} \\
 & \Rightarrow S=\dfrac{\left( \sqrt{3}+2 \right)K}{2} \\
\end{align} $
We know that the side opposite to the greatest angle of a triangle is the greatest side. So, In triangle ABC, angle A is the greatest, which implies that a is the greatest side. So, the ratio of S to a will be given by,
 $ \begin{align}
  & \dfrac{S}{a}=\dfrac{\left( \dfrac{\sqrt{3}+2}{2} \right)K}{\left( \dfrac{\sqrt{3}}{2} \right)K} \\
 & \Rightarrow \dfrac{S}{a}=\dfrac{\sqrt{3}+2}{\sqrt{3}} \\
 & \Rightarrow S:a=\sqrt{3}+2:\sqrt{3} \\
\end{align} $
Therefore, the correct option is option A.

Note: The students may sometimes take the value of $ \sin 120{}^\circ =\dfrac{1}{2} $ by mistake, which will give an incorrect value of a and thus we will get the incorrect answer. We should remember that the value of $ \sin 120{}^\circ =\dfrac{\sqrt{3}}{2} $ . It is also important to know the largest side of the triangle and also that the side opposite to the largest angle is the largest side of the triangle. We should use the sine rule for the triangles carefully and it states that the ratio of side to the angles opposite to it, is always constant.