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

Hint- Whenever we talk about three unknown numbers in arithmetic progression always take numbers as $(a - d),{\text{ }}a,{\text{ }}(a + d)$ where $d$ is the common difference. Here assume angles in the same way and then apply the angle sum property of triangles.

Complete step-by-step answer:
Three angles in A.P.
If $y$ is a common difference let these angles be ${(x - y)^0},{x^0}and{(x + y)^0}$
$
   \Rightarrow x - y + x + x + y = {180^0}{\text{ }}\left[ {\because {\text{angle sum property}}} \right] \\
  3x = {180^0} \\
  x = \dfrac{{{{180}^0}}}{3} = {60^0} \\
$
Now, the greatest angle
$
   = {(x + y)^0} \\
   = (x + y)\dfrac{\pi }{{180}}{\text{ }}radians \\
$
According to question
$
  \dfrac{{(x - y)}}{{(x + y) \times \dfrac{\pi }{{180}}}} = \dfrac{{60}}{\pi } \\
  (x - y) = (x + y) \times \dfrac{\pi }{{180}} \times \dfrac{{60}}{\pi } \\
  (x - y) = (x + y) \times \dfrac{1}{3} \\
  3x - 3y = x + y \\
  {\text{rearranging terms}} \\
  3y + y = 3x - x \\
  4y = 2x \\
  y = \dfrac{{2x}}{4} \\
  y = \dfrac{x}{2} \\
  {\text{put value of x = 6}}{{\text{0}}^0} \\
  y = \dfrac{{{{60}^0}}}{2} = {30^0} \\
$

Therefore, three angles in A.P. are
$
   = (x - y),x,(x + y) \\
   = {30^0},{60^0},{90^0} \\
$
Hence, the smallest angles in degrees is ${30^0}$

Note- For this type of question properties of arithmetic progression should be known and also the angle properties of triangles and one should also know the conversion of degree to radian and vice-versa.