Question

In $\Delta ABC$ if $\angle C=3\angle B$ and $\angle B=\dfrac{2}{3}\left[ \angle A+\angle B \right]$, find the values of $\angle A,\angle B,\angle C$ respectively.

Answer 1

Hint: We know that the sum of angles in a triangle is equal to $180{}^\circ $. From the given data we will convert the three angles into one variable and find the sum of the angles in the triangle and equate them to $180{}^\circ $ . Then we will get the value of that variable and using that value we will find the angles of the triangle.

Complete step-by-step answer:
Given that, In $\Delta ABC$ ,$\angle C=3\angle B$ and $\angle B=\dfrac{2}{3}\left[ \angle A+\angle B \right]$
The $\Delta ABC$ is represented in below picture

Let the angles in the above triangle are
$\angle CAB=\alpha $
$\angle ABC=\beta $
$\angle ACB=\gamma $
From the given data
$\begin{align}
& \angle ABC=\beta =\dfrac{2}{3}\left[ \angle CAB+\angle ABC \right] \\
& \beta =\dfrac{2}{3}\left[ \alpha +\beta \right] \\
& \beta -\dfrac{2}{3}\beta =\dfrac{2}{3}\alpha \\
& \dfrac{1}{3}\beta =\dfrac{2}{3}\alpha
\end{align}$
Multiplying the above equation with $3$, then
$\begin{align}
 & \beta =2\alpha \\
 & \alpha =\dfrac{1}{2}\beta .....\left( \text{i} \right)
\end{align}$
Now given that
$\begin{align}
 & \angle ACB=3\angle ABC \\
 & \gamma =3\beta ......\left( \text{ii} \right)
\end{align}$
Finding the sum of the angles in the triangle,
$\angle CAB+\angle ABC+\angle ACB=\alpha +\beta +\gamma $
From equations $\left( \text{i} \right)$ and $\left( \text{ii} \right)$ substituting the values of $\alpha ,\gamma $ in the above equation, then
$\begin{align}
 & \angle CAB+\angle ABC+\angle ACB=\alpha +\beta +\gamma \\
 & =\dfrac{1}{2}\beta +\beta +3\beta \\
 & \angle CAB+\angle ABC+\angle ACB=\dfrac{9}{2}\beta .......\left( \text{iii} \right)
\end{align}$
But we know that the sum of angles in a triangle is equal to $180{}^\circ $, so
$\angle CAB+\angle ABC+\angle ACB=180{}^\circ $
From equation $\left( \text{iii} \right)$ substituting the value of $\angle CAB+\angle ABC+\angle ACB$ in the above equation, then
$\begin{align}
 & \dfrac{9}{2}\beta =180{}^\circ \\
 & 9\beta =360{}^\circ \\
 & \beta =\dfrac{360{}^\circ }{9} \\
 & =40{}^\circ
\end{align}$
From equation $\left( \text{i} \right)$ the value of $\alpha $ is
$\begin{align}
 & \alpha =\dfrac{1}{2}\beta \\
 & =\dfrac{1}{2}\left( 40{}^\circ \right) \\
 & =20{}^\circ
\end{align}$
From equation $\left( \text{ii} \right)$ the value of $\gamma $ is
$\begin{align}
 & \gamma =3\beta \\
 & =3\left( 40{}^\circ \right) \\
 & =120{}^\circ
\end{align}$
So, the angles of the triangle are
$\begin{align}
 & \angle CAB=\angle A =\alpha \\
 & \angle A=20{}^\circ
\end{align}$
$\begin{align}
 & \angle ABC=\angle B =\beta \\
 & \angle B=40{}^\circ
\end{align}$
$\begin{align}
 & \angle ACB=\angle C=\gamma \\
 & \angle C=120{}^\circ
\end{align}$

Note: We can convert all the variables to any one of the variables either $\alpha $ or $\beta $ or $\gamma $ according to the given statement. From equation $\left( \text{i} \right)$ we can write $\beta =2\alpha $, now substitute the value of $\beta $ in equation $\left( \text{ii} \right)$, then we will get
$\begin{align}
  & \gamma =3\beta \\
 & =3\left( 2\alpha \right) \\
 & =6\alpha
\end{align}$
Now here we can write
$\begin{align}
  & \angle CAB+\angle ABC+\angle ACB=\alpha +\beta +\gamma \\
 & =\alpha +2\alpha +6\alpha \\
 & =9\alpha
\end{align}$
Equating the value of $\angle CAB+\angle ABC+\angle ACB$ to $180{}^\circ $, then we have
$\begin{align}
  & 9\alpha =180{}^\circ \\
 & \alpha =20{}^\circ
\end{align}$
Here from both methods we get the same answers. So, there is no restriction to convert all the angles into $\beta $.