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

Let the angles of the triangle be $\left(a-d\right)°,\left(a\right)°\mathrm{and}\left(a+d\right)°$.
We know:
$$\begin{gathered} a-d+a+a+d=180 \\ \Rightarrow 3a=180 \\ \Rightarrow a=60 \end{gathered}$$
Given:
$$\begin{gathered} \mathrm{Greatest}\mathrm{angle}=5\times \mathrm{Least}\mathrm{angle} \\ \mathrm{or},\frac{\mathrm{Greatest}\mathrm{angle}}{\mathrm{Least}\mathrm{angle}}=5 \\ \mathrm{or},\frac{a+d}{a-d}=5 \\ \mathrm{or},\frac{60+d}{60-d}=5 \\ \mathrm{or},60+d=300-5d \\ \mathrm{or},6d=240 \\ \mathrm{or},d=40 \end{gathered}$$

Hence, the angles are $\left(a-d\right)°,\left(a\right)°\mathrm{and}\left(a+d\right)°$, i.e., $20°,60°\mathrm{and}100°$, respectively.

∴ Angles of the triangle in radians = $\left(20\times {\displaystyle \frac{\mathrm{\pi }}{180}}\right),\left(60\times {\displaystyle \frac{\mathrm{\pi }}{180}}\right)\mathrm{and}\left(100\times {\displaystyle \frac{\mathrm{\pi }}{180}}\right)$
=$\frac{\mathrm{\pi }}{9},\frac{\mathrm{\pi }}{3}\mathrm{and}\frac{5\mathrm{\pi }}{9}$