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 \(1:2\). Find the angles in radians.
Solution:
Let the three angles in A.P. be:
\[ a-d,\ a,\ a+d \]
Given,
\[ \frac{a-d}{a}=\frac{1}{2} \]
\[ 2a-2d=a \]
\[ a=2d \]
Sum of angles of a triangle:
\[ (a-d)+a+(a+d)=180^\circ \]
\[ 3a=180^\circ \]
\[ a=60^\circ \]
Since,
\[ a=2d \]
\[ 60^\circ=2d \]
\[ d=30^\circ \]
Therefore, the angles are:
\[ 30^\circ,\ 60^\circ,\ 90^\circ \]
Convert into radians:
\[ 30^\circ=\frac{\pi}{6} \]
\[ 60^\circ=\frac{\pi}{3} \]
\[ 90^\circ=\frac{\pi}{2} \]
Hence, the angles in radians are:
\[ \frac{\pi}{6},\ \frac{\pi}{3},\ \frac{\pi}{2} \]