Lines AB and CD intersect at P. If angle APC | vedclass

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Question

(A) When two lines intersect,the vertically opposite angles are equal.Since lines $AB$ and $CD$ intersect at $P$,$\angle APC$ and $\angle BPD$ are ver

Vedclass · Accepted Answer

$x = 25, \angle APC = 80^{\circ}, \angle BPD = 80^{\circ}$

Vedclass · Answer

(A) When two lines intersect,the vertically opposite angles are equal.Since lines $AB$ and $CD$ intersect at $P$,$\angle APC$ and $\angle BPD$ are vertically opposite angles.Therefore,$\angle APC = \angle BPD$.Substituting the given expressions: $2x + 30^{\circ} = 4x - 20^{\circ}$.Rearranging the terms: $30^{\circ} + 20^{\circ} = 4x - 2x$.$50^{\circ} = 2x$,which gives $x = 25^{\circ}$.Now,calculate the angles:$\angle APC = 2(25) + 30 = 50 + 30 = 80^{\circ}$.$\angle BPD = 4(25) - 20 = 100 - 20 = 80^{\circ}$.Thus,$x = 25$,$\angle APC = 80^{\circ}$,and $\angle BPD = 80^{\circ}$.

Lines $AB$ and $CD$ intersect at $P$. If $\angle APC = 2x + 30^{\circ}$ and $\angle BPD = 4x - 20^{\circ}$,then find $x$,$\angle APC$,and $\angle BPD$.

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