# The angles of a quadrilateral are in A.P. whose common difference is 10. Find the angles.

$Given:common\;difference=10\\*We\;know\;that\;the\;sum\;of\;all\;angles\;in\;a\;quadrilateral\;is\;360^{\circ}\\*Let\;us\;assume\;the\;angles\;are\;a-3d, a-d,a+d,a+3d\\*So,\;a-3d+a-d+a+d+a+3d=360^{\circ}\\* \Rightarrow 4a=360^{\circ}\\*\Rightarrow a=90\\*And,\\*\Rightarrow common\;difference=(a-d)-(a-3d)=10\\*\Rightarrow 2d=10\\*\Rightarrow d=\frac{10}{2}=5\\*Hence,\;the\;angles\;are\;a-3d,\;a-d,\;a+d,\;a+3d \; which\;is\;75^{\circ},\;85^{\circ},\;95^{\circ},\;105^{\circ}$

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