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

Answer: (C)

$\frac{\pi }{3}$

Let the four angles of a quadrilateral are
$(a-3d{)}^{\circ }$, $(a-d{)}^{\circ }$, $(a+d{)}^{\circ }$ and $(a+3d{)}^{\circ }$
$\therefore a-3d+a-d+a+d+a+3d={360}^{\circ }⟹a={90}^{\circ }$
But the greatest angle is double the least, i.e.,
$a+3d=2\times (a-3d)=2a-6d$
$\therefore 9d=2a-a=a={90}^{\circ }$

$\therefore d=\frac{90}{9}={10}^{\circ }$

$\therefore$ Least angle $a-3d={90}^{\circ }-3\times {10}^{\circ }$

$={60}^{\circ }=\frac{\pi }{3}$