Question

The angles of a triangle are in A.P. and the number of degrees in the least is to the number of radians in the greatest as 60 to ${\pi }^c$. Find the smallest angle in degrees.

Answer 1

The three angles in A.P.;
if y is common difference, let these angles be ${(x+y)}^{\circ },x^{\circ }and{(x-y)}^{\circ }$
$\therefore x+y+x+x-y={180}^{\circ }$
$x={60}^{\circ }$
According to the question,
$\frac{(x-y)}{(x+y)\frac{{\pi }^c}{180}}=\frac{60}{\pi }$
$\pi (x-y)=(x+y)\frac{\pi }{180}\times 60$
$3(x-y)=x+y$
$4y=2x$
$y=\frac{x}{2}$
$\therefore y=\frac{{60}^{\circ }}{2}={30}^{\circ }$
Hence three angles are ${30}^{\circ },{60}^{\circ }and{90}^{\circ }.$