Question

The angles of a triangle are in A.P. and the number of degrees in the mean angle is 1: 120. Find the angles in radians.

Answer 1

Hint: First of all we will suppose the angles in A.P. as a, a+d, a+2d and then we will use the property of a triangle that the sum of interior angles of a triangle is equal to \[{{180}^{\circ }}\].

Complete step-by-step answer:
Let us suppose the first term and common difference to be ‘a’ and ‘d’ respectively of the angles which are in A.P.
So the angles are a, a+d, a+2d.
We know that \[{{T}_{n}}\] \[=a+(n-1)d\] for an A.P.
We know that the sum of the interior angles of a triangle is \[{{180}^{\circ }}\].
\[\begin{align}
  & \Rightarrow a+\left( a+d \right)+\left( a+2d \right)={{180}^{\circ }} \\
 & \Rightarrow a+a+d+a+2d={{180}^{\circ }} \\
\end{align}\]
On adding the similar terms to the left hand side of the equation we get as follows:
\[\Rightarrow 3a+3d={{180}^{\circ }}\]
On taking 3 as common, we get as follows:
\[\Rightarrow 3\left( a+d \right)=3({{60}^{\circ }})\]
On dividing the equation by 3, we get as follows:
\[\begin{align}
  & \Rightarrow \dfrac{3\left( a+d \right)}{3}=\dfrac{3}{3}({{60}^{\circ }}) \\
 & \Rightarrow \left( a+d \right)={{60}^{\circ }}......(1) \\
\end{align}\]
We have been given that the number of degrees in the mean angle is 1:120.
\[\Rightarrow \dfrac{a}{a+d}=\dfrac{1}{120}\]
On cross multiplication, we get as follows:
\[\Rightarrow 120a=a+d\]
On taking a to the left hand side, we get as follows:
\[\begin{align}
  & \Rightarrow 120a-a=d \\
 & \Rightarrow 119a=d \\
\end{align}\]
On substituting the value of \[d={{60}^{\circ }}-a\] from equation (1) we get as follows:
\[\begin{align}
  & \Rightarrow 119a=60-a \\
 & \Rightarrow 119a+a=60 \\
 & \Rightarrow 120a=60 \\
 & \Rightarrow \dfrac{120a}{120}=\dfrac{60}{120} \\
 & \Rightarrow a={{\dfrac{1}{2}}^{\circ }} \\
\end{align}\]
On substituting \[a={{\dfrac{1}{2}}^{\circ }}\] in equation (1) we get as follows:
\[\begin{align}
  & \Rightarrow \dfrac{1}{2}+d={{60}^{\circ }} \\
 & \Rightarrow d={{60}^{\circ }}-\dfrac{1}{2} \\
 & \Rightarrow d=\dfrac{120-1}{2} \\
 & \Rightarrow d=\dfrac{{{119}^{\circ }}}{2} \\
\end{align}\]
The angles in degrees are as follows:
\[a={{\dfrac{1}{2}}^{\circ }},a+d=\dfrac{1}{2}+\dfrac{119}{2}={{60}^{\circ }},a+2d=\dfrac{1}{2}+2\left( \dfrac{119}{2} \right)=\dfrac{1}{2}+119={{\dfrac{239}{2}}^{\circ }}\]
Now as we know that 1 degree \[\dfrac{\pi }{180}\] radians
\[\begin{align}
  & \Rightarrow {{\dfrac{1}{2}}^{\circ }}=\dfrac{\pi }{180}\times \dfrac{1}{2}=\dfrac{\pi }{360}\text{radians} \\
 & \Rightarrow {{60}^{\circ }}=\dfrac{\pi }{180}\times {{60}^{\circ }}=\dfrac{\pi }{3}\text{radians} \\
 & \Rightarrow \dfrac{239}{2}=\dfrac{\pi }{180}\times \dfrac{239}{2}=\dfrac{239\pi }{360}\text{radians} \\
\end{align}\]
Therefore the angles in radians are \[\dfrac{\pi }{360}\text{radians},\dfrac{\pi }{3}\text{radians},\dfrac{239\pi }{360}\text{radians} \].

Note: Be careful of the sign while doing calculation and also while conversion of degree into radians, use the correct formula properly, don't use 1 degree \[=\dfrac{180}{\pi }\] radians in a hurry. Also, remember that the sum of all internal angles of a triangle is equal to \[180\] degree or \[\pi \] radians. Also we can assume the terms of A.P. as a-d, a, a+d the simplification becomes easy if we suppose these terms.