Question

The angles of a quadrilateral are in A.P and the greatest is double the least, Find the least angles 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 four angles of a quadrilateral are
$(a-3d{)}^{\circ },(a-d{)}^{\circ },(a+d{)}^{\circ }and(a+3d{)}^{\circ }$
$\therefore a-3d+a-d+a+d+a+3d={360}^{\circ }$
$\therefore a={90}^{\circ }$
But the greatest angle is double the least , i.e.
$a+3d=2\times (a-3d)=2a-6d$
$\therefore 9a=2a-a=a={90}^{\circ }$
$\therefore d=\frac{90}{9}={10}^{\circ }$
therefore Least angle =a-3d
$={90}^{\circ }-3\times {10}^{\circ }={60}^{\circ }=\frac{\pi }{3}$