Question

The angles of a triangle are in A.P. such that the greatest is 5 times the least. Find the angles in radians.

Answer 1

Hint: First of all we will assume the angles in A.P. with first term and common difference as a variable, say a and d. Also, we will use the conversion of degree into radians as follows:
1 degree \[=\dfrac{\pi }{180}\] radians
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’ of the angles of triangle 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 all interior angles of a triangle is equal to \[{{180}^{\circ }}\].
\[\begin{align}
  & \Rightarrow a+\left( a+d \right)+\left( a+2d \right)={{180}^{\circ }} \\
 & \Rightarrow a+a+d+a+2d={{180}^{\circ }} \\
\end{align}\]
On adding the similar terms to left hand side of equality 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:
\[\begin{align}
  & \Rightarrow \dfrac{3\left( a+d \right)}{3}=\dfrac{3}{3}\left( {{60}^{\circ }} \right) \\
 & \Rightarrow \left( a+d \right)={{60}^{\circ }}......(1) \\
\end{align}\]
We have been given that the greatest angle is 5 times the least.
Now ‘a’ is least and ‘a+2d’ is greatest.
\[\begin{align}
  & \Rightarrow a+2d=5\times a \\
 & \Rightarrow a+2d=5a \\
 & \Rightarrow 2d=5a-a \\
 & \Rightarrow 2d=4a \\
 & \Rightarrow d=\dfrac{4a}{2}=2a......(2) \\
\end{align}\]
Now substituting d = 2d in equation (1) we get as follows:
\[\begin{align}
  & \Rightarrow a+2a={{60}^{\circ }} \\
 & \Rightarrow 3a={{60}^{\circ }} \\
 & \Rightarrow a=\dfrac{{{60}^{\circ }}}{3}={{20}^{\circ }} \\
\end{align}\]
Now substituting \[a={{20}^{\circ }}\] in equation (2) we get as follows:
\[d=2\times {{20}^{\circ }}={{40}^{\circ }}\]
Hence, the angle in degree is as follows:
\[a={{20}^{\circ }},a+d=20+40={{60}^{\circ }},a+2d=20+2\times 40={{100}^{\circ }}\]
Now we know that 1 degree \[=\dfrac{\pi }{180}\] radians
 \[\begin{align}
  & \Rightarrow {{20}^{\circ }}=\dfrac{\pi }{180}\times 20=\dfrac{\pi }{9}\text{radians} \\
 & \Rightarrow {{60}^{\circ }}=\dfrac{\pi }{180}\times 60=\dfrac{\pi }{3}\text{radians}\\
 & \Rightarrow {{100}^{\circ }}=\dfrac{\pi }{180}\times 100=\dfrac{5\pi }{9}\text{radians} \\
\end{align}\]
Therefore, the angles in radians are \[\dfrac{\pi }{9}\text{radians},\dfrac{\pi }{3}\text{radians},\dfrac{5\pi }{9}\text{radians}\].

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