Question

The ratio of the sides of the triangle $\vartriangle ABC$ _{is $1:\sqrt{3}:2$.} The ratio of $A: B:C$ _is
A. $3:5:2$
B. $1:\sqrt{3}:2$
C. $3:2:1$
D. $1:2:3$

Answer 1

Hint: To solve this question we will take the ratio of the lengths of the sides of the triangle and then derive the value of $a,b$and $c$. We will then substitute the values of the lengths in the three cosine rules with different angles and find the value of the angles $A,B$ and $C$ . Then we will determine their ratio.
Formula used:
There are three cosine rule or Law of cosine with different angles.
\[{{c}^{2}}={{a}^{2}}+{{b}^{2}}-2ab\cos C\]
\[{{a}^{2}}={{b}^{2}}+{{c}^{2}}-2bc\cos A\]
\[{{b}^{2}}={{a}^{2}}+{{c}^{2}}-2ac\cos B\]
Complete step-by-step solution:
We are given a triangle $\vartriangle ABC$having ratio of sides $1:\sqrt{3}:2$_{and we have to find the ratio of $A:B:C$}
Let the length of sides of the triangle be $x$, So the value of lengths will be ,
$\begin{align}
  & a=x \\
 & b=\sqrt{3}x \\
 & c=2x \\
\end{align}$
We will now substitute the values of $a,b$and $c$in the cosine rule \[{{c}^{2}}={{a}^{2}}+{{b}^{2}}-2ab\cos C\]and find the value of angle $C$.
\[\begin{align}
  & {{(2x)}^{2}}={{x}^{2}}+{{\left( \sqrt{3}x \right)}^{2}}-2x(\sqrt{3}x)\cos C \\
 & 4{{x}^{2}}={{x}^{2}}+3{{x}^{2}}-2\sqrt{3}{{x}^{2}}\cos C \\
 & 0=-2\sqrt{3}{{x}^{2}}\cos C \\
 & \cos C=0 \\
 & \cos C=\cos {{90}^{0}} \\
 & C={{90}^{0}}
\end{align}\]
We will again substitute the values of $a,b$and $c$in the cosine rule \[{{a}^{2}}={{b}^{2}}+{{c}^{2}}-2bc\cos A\]and find the value of angle $A$.
\[\begin{align}
  & {{x}^{2}}={{(\sqrt{3}x)}^{2}}+{{(2x)}^{2}}-2(\sqrt{3}x)2x\cos A \\
 & {{x}^{2}}=3{{x}^{2}}+4{{x}^{2}}-4\sqrt{3}{{x}^{2}}\cos A \\
 & -6{{x}^{2}}=-4\sqrt{3}{{x}^{2}}\cos A \\
 & \cos A=\frac{\sqrt{3}}{2} \\
 & \cos A=\cos {{30}^{0}} \\
 & A={{30}^{0}}
\end{align}\]
We will again substitute the values of $a,b$and $c$in the cosine rule \[{{b}^{2}}={{a}^{2}}+{{c}^{2}}-2ac\cos B\]and find the value of angle $B$.
\[\begin{align}
  & {{(\sqrt{3}x)}^{2}}={{x}^{2}}+{{(2x)}^{2}}-2x(2x)\cos B \\
 & 3{{x}^{2}}={{x}^{2}}+4{{x}^{2}}-4{{x}^{2}}\cos B \\
 & -2{{x}^{2}}=-4{{x}^{2}}\cos B \\
 & \frac{1}{2}=\cos B \\
 & \cos {{60}^{0}}=\cos B \\
 & B={{60}^{0}}
\end{align}\]
Now the ratio of the angles will be,
$\begin{align}
  & A:B:C=30:60:90 \\
 & A:B:C=1:2:3 \\
\end{align}$
The ratio of the angles $A:B:C$of the triangle $\vartriangle ABC$_{such that the ratio of their sides are $1:\sqrt{3}:2$ is equals to $A:B:C=1:2:3$}.Hence the correct option is (D).
Note:
If you know the other two sides of a triangle and the angle between them, the law of cosines, often known as the "cosine law," instructs you on how to find the third side. You cannot accomplish it because of the sine law, sometimes known as the law of sines. As an alternative, it informs you that the sines of the angles are inversely proportional to the lengths of the sides that they are opposing. If you know two sides and the angles across from those two sides, for instance, you could use it. To find angle $B$, we could have also used angle sum property of the triangle according to which sum of all the angles in a triangle is equals to ${{180}^{0}}$.
$\begin{align}
  & A+B+C={{180}^{0}} \\
 & {{30}^{0}}+B+{{90}^{0}}={{180}^{0}} \\
 & B={{60}^{0}}
\end{align}$