Question

The angles of a quadrilateral are in A.P. and the greatest angle is double the least; express the angles in radians.

Answer 1

Let the angles be $\theta -3d,\theta -d,\theta +d,\theta +3d$

which are in $A.P$ with $2d$ as common difference

According to :-

$\therefore$ Given condition

$\theta +3d=2(\theta -3d)$ ...(1)

& $(\theta -3d)+(\theta -d)+(\theta +d)+(\theta +3d)=360$ ...(2)

$\therefore 4\theta =360$

$\therefore \theta =90$

put $\theta ={90}^o$ in equation (1)

$\Rightarrow 90+3d=2(90-3d)$

$\Rightarrow 9d=90\Rightarrow d=10$

Hence Ls are :-

$60,80,100,120$

Angles in Radius :- $\frac{\pi }{3},\frac{80\times \pi }{180},\frac{100\times \pi }{180},\frac{120\times \pi }{180}$