If all interior angles of a quadrilateral are | vedclass

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(D) Let the four angles of the quadrilateral in $A.P.$ be $(a-3d), (a-d), (a+d), (a+3d)$.The common difference of this sequence is $2d = 10^{\circ}$,w

Vedclass · Accepted Answer

$75$

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(D) Let the four angles of the quadrilateral in $A.P.$ be $(a-3d), (a-d), (a+d), (a+3d)$.The common difference of this sequence is $2d = 10^{\circ}$,which implies $d = 5^{\circ}$.The sum of interior angles of a quadrilateral is $360^{\circ}$.$(a-3d) + (a-d) + (a+d) + (a+3d) = 360^{\circ}$$4a = 360^{\circ} \Rightarrow a = 90^{\circ}$.The smallest angle is $a-3d = 90^{\circ} - 3(5^{\circ}) = 90^{\circ} - 15^{\circ} = 75^{\circ}$.

If all interior angles of a quadrilateral are in $A.P.$ and the common difference is $10^{\circ}$,find the smallest angle. (in $^{\circ}$)

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