The angles of a triangle are in A.P. such that the greatest angle is 5 times the least. Find the angles in radians.
Solution:
Let the angles in A.P. be:
\[ a-d,\ a,\ a+d \]
Given,
\[ a+d=5(a-d) \]
\[ a+d=5a-5d \]
\[ 6d=4a \]
\[ 3d=2a \]
Sum of angles of a triangle:
\[ (a-d)+a+(a+d)=180^\circ \]
\[ 3a=180^\circ \]
\[ a=60^\circ \]
Now,
\[ 3d=2a \]
\[ 3d=120^\circ \]
\[ d=40^\circ \]
Therefore, the angles are:
\[ 20^\circ,\ 60^\circ,\ 100^\circ \]
Convert into radians:
\[ 20^\circ=\frac{\pi}{9} \]
\[ 60^\circ=\frac{\pi}{3} \]
\[ 100^\circ=\frac{5\pi}{9} \]
Hence, the angles in radians are:
\[ \frac{\pi}{9},\ \frac{\pi}{3},\ \frac{5\pi}{9} \]