The measures of all the four angles of a quad | vedclass

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(B) Let the four angles of the quadrilateral be $a-3d, a-d, a+d, a+3d$. Here,the common difference between consecutive terms is $2d = 10^{\circ}$,so $

Vedclass · Accepted Answer

$75^{\circ}, 85^{\circ}, 95^{\circ}, 105^{\circ}$

Vedclass · Answer

(B) Let the four angles of the quadrilateral be $a-3d, a-d, a+d, a+3d$. Here,the common difference between consecutive terms is $2d = 10^{\circ}$,so $d = 5^{\circ}$. The sum of the angles of a quadrilateral is $360^{\circ}$. Therefore,$(a-3d) + (a-d) + (a+d) + (a+3d) = 360^{\circ}$. $4a = 360^{\circ} \implies a = 90^{\circ}$. The angles are: $a-3d = 90^{\circ} - 3(5^{\circ}) = 90^{\circ} - 15^{\circ} = 75^{\circ}$. $a-d = 90^{\circ} - 5^{\circ} = 85^{\circ}$. $a+d = 90^{\circ} + 5^{\circ} = 95^{\circ}$. $a+3d = 90^{\circ} + 3(5^{\circ}) = 90^{\circ} + 15^{\circ} = 105^{\circ}$. Thus,the angles are $75^{\circ}, 85^{\circ}, 95^{\circ}, 105^{\circ}$.

The measures of all the four angles of a quadrilateral form an $A.P.$. If the common difference is $10^{\circ}$,find the measures of all the angles.

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