Question

The angles of a quadrilateral are in A.P. and the greatest is double the least. Find the least angle in radian.

• A. $\frac{\pi }{6}$
• B. $\frac{\pi }{4}$
• C. $\frac{\pi }{3}$
• D. $\frac{\pi }{5}$

Answer 1

The correct option is C $\frac{\pi }{3}$

Let the angles be $(a-3d{)}^0,(a-d{)}^0,(a+d{)}^0,(a+3d{)}^0$ being the common difference of the A.P.
Therefore, sum of all the angles of a quadrilateral is always ${360}^0$.
$(a-3d{)}^0+(a-d{)}^0+(a+d{)}^0+(a+3d{)}^0={360}^0$
$4a={360}^0$
$a={90}^0$ .....(i)
It is given that the greatest angle is twice of the least
Therefore, $a+3d=2(a-3d)$
$\Rightarrow a=9d$
Substituting the value of $a$ from (i)
${90}^0=9d$
$d={10}^0$
Hence the least angle is $(a-3d{)}^0={90}^0-{30}^0={60}^0$ $=\frac{\pi }{3}$
Hence, the answer is C.