Question

The angles of a quadrilateral are in A.P. and the greatest angle is double the least angle. Find the least angles in radians.
A. \[\dfrac{\pi }{6}\]
B. \[\dfrac{\pi }{4}\]
C. \[\dfrac{\pi }{3}\]
D. \[\dfrac{\pi }{5}\]

Answer 1

Hint: First of all, consider the angles of the quadrilateral as \[{\left( {a - 3d} \right)^0},{\left( {a - d} \right)^0},{\left( {a + d} \right)^0}{\text{ and }}{\left( {a + 3d} \right)^0}\] since they are in A.P. Then equate their sum to \[{360^0}\]. Then find the least angle of the quadrilateral by using the given conditions.

Complete step-by-step answer:

Given the angles of a quadrilateral are in A.P.
Let the angles be \[{\left( {a - 3d} \right)^0},{\left( {a - d} \right)^0},{\left( {a + d} \right)^0}{\text{ and }}{\left( {a + 3d} \right)^0}\]
We know that the sum of the angles in a quadrilateral is equal to \[{360^0}\]. Then we have
\[
   \Rightarrow a - 3d + a - d + a + d + a + 3d = 360 \\
   \Rightarrow 4a = 360 \\
  \therefore a = 90 \\
\]

But given that the greatest angle is equal to double of the least angle. So, we have
\[
   \Rightarrow a + 3d = 2\left( {a - 3d} \right) \\
   \Rightarrow a + 3d = 2a - 6d \\
   \Rightarrow 6d + 3d = 2a - a \\
   \Rightarrow 9d = a \\
   \Rightarrow 9d = 90{\text{ }}\left[ {\because a = 90} \right] \\
  \therefore d = 10 \\
\]

Therefore, the least angle is \[{\left( {a - 3d} \right)^0} = \left( {90 - 3 \times 10} \right) = {60^0}\]

By using the conversion \[\pi = {180^0}\], we have
\[{\left( {a - 3d} \right)^0} = {60^0} = \dfrac{\pi }{3}\]
Thus, the correct option is C. \[\dfrac{\pi }{3}\]

Note: The sum of the angles in a quadrilateral is equal to \[{360^0}\]. To convert degrees to radians, take the number of degrees to be converted and multiply it by \[\dfrac{\pi }{{180}}\]. The difference in the angles of the quadrilateral are equal as they are in A.P.