Question

The angles of a triangle are in A.P. and the number of degrees in the least angle is to the number of degrees in the mean angle as 1 : 120. 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:
$$\begin{gathered} a-d+a+a+d=180 \\ \Rightarrow 3a=180 \\ \Rightarrow a=60 \end{gathered}$$

$$\begin{gathered} \mathrm{Given}: \\ \frac{\mathrm{Number}\mathrm{of}\mathrm{degrees}\mathrm{in}\mathrm{the}\mathrm{least}\mathrm{angle}}{\mathrm{Number}\mathrm{of}\mathrm{degrees}\mathrm{in}\mathrm{the}\mathrm{mean}\mathrm{angle}}=\frac{1}{120} \\ \mathrm{or},\frac{a-d}{a}=\frac{1}{120} \\ \mathrm{or},\frac{60-d}{60}=\frac{1}{120} \\ \mathrm{or},\frac{60-\mathrm{d}}{1}=\frac{1}{2} \\ \mathrm{or},120-2d=1 \\ \mathrm{or},2\mathrm{d}=119 \\ \mathrm{or},d=59.5 \end{gathered}$$

Hence, the angles are $\left(a-d\right)°,\left(a\right)°\mathrm{and}\left(a+d\right)°$, i.e., $0.5°,60°\mathrm{and}119.5°$.

∴ Angles of the triangle in radians = $\left(0.5\times {\displaystyle \frac{\mathrm{\pi }}{180}}\right),\left(60\times {\displaystyle \frac{\mathrm{\pi }}{180}}\right)\mathrm{and}\left(119.5\times {\displaystyle \frac{\mathrm{\pi }}{180}}\right)$
= $\frac{\mathrm{\pi }}{360},\frac{\mathrm{\pi }}{3}\mathrm{and}\frac{239\mathrm{\pi }}{360}$