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
\[\left( a - d \right)^\circ, \left( a \right)^\circ \text{ and } \left( a + d \right)^\circ\].
We know:
\[a - d + a + a + d = 180\]
\[ \Rightarrow 3a = 180\]
\[ \Rightarrow a = 60\]
Given:
\[\frac{\text{ Number of degrees in the least angle }}{\text{ Number of degrees in the mean angle }} = \frac{1}{120}\]
\[\text{ or, } \frac{a - d}{a} = \frac{1}{120}\]
\[\text{ or, }\frac{60 - d}{60} = \frac{1}{120}\]
\[\text{ or, }\frac{60 - d}{1} = \frac{1}{2}\]
\[\text{ or,} 120 - 2d = 1\]
\[\text{ or,} 2d = 119\]
\[\text{ or,} d = 59 . 5\]
Hence, the angles are
\[\left( a - d \right)^\circ, \left( a \right)^\circ \text{ and }\left( a + d \right)^\circ\]

\[0 . 5^\circ, 60^\circ\text{ and }119 . 5^\circ\]
∴ Angles of the triangle in radians = \[\left( 0 . 5 \times \frac{\pi}{180} \right), \left( 60 \times \frac{\pi}{180} \right)\text{ and }\left( 119 . 5 \times \frac{\pi}{180} \right)\]

\[\frac{\pi}{360}, \frac{\pi}{3}\text{ and }\frac{239\pi}{360}\]