Question

The angles of a quadrilateral are in AP, and the greatest angle is double the least. Express the least angle in radians.

[FULL SOLUTION REQUIRED]

[CONTENT QUALITY IS MUST]

Answer 1

solution

let the smallest term be x

and, largest term be 2x

Then AP formed= x,?,?, 2x

so, Sn= n/2 [2a+ (n-1)d]

=> sn= n/2 [a+ a(n-1)d]

=> 360°= 4/2 [x+ 2x]....[We know that → a+(n-1) d= last term= 2x]

=> 180°= 3x

=> x= 60°

Now, 60° is least angle.

= 60°= π/180° *60°

=> 60° = π/3 rad

________________
Nice question.

hope you get it,☺☺

Answer 2

Hey mate!

Here's your answer!!

Let the least angle be x. And the common difference is D. Then the four angles will be
x, x+d, x+2d, x+3d.

Now by property of quadrilaterals, sum of all angles is 360 degrees, therefore,
x+ x+d+ x+2d+ x+3d = 360
=> 4x + 6d = 360 (divide equation by 2)
=> 2x + 3d = 180

By the given condition,
x+3d (greatest angle) = 2 * x(smallest angle)

=> x + 3d = 2x
=> 3d = x
put this in above equation,

2x + x = 180
=> x = 60 degrees.

Now solving equation 3d = x by putting x = 60
d = 20

•Thus the four angles are
60, 80, 100, 120.

$hope \: it \: helps \: you$
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