angle APC and angle BPD are vertically opposi | vedclass

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Question

(A) Since $\angle APC$ and $\angle BPD$ are vertically opposite angles,they must be equal.Therefore,$3x - 20^{\circ} = 2x + 30^{\circ}$.Subtracting $2

Vedclass · Accepted Answer

$x = 50^{\circ}, 130^{\circ}, 130^{\circ}$

Vedclass · Answer

(A) Since $\angle APC$ and $\angle BPD$ are vertically opposite angles,they must be equal.Therefore,$3x - 20^{\circ} = 2x + 30^{\circ}$.Subtracting $2x$ from both sides,we get $x - 20^{\circ} = 30^{\circ}$.Adding $20^{\circ}$ to both sides,we get $x = 50^{\circ}$.Now,substitute $x = 50^{\circ}$ into the expressions for the angles:$\angle APC = 3(50^{\circ}) - 20^{\circ} = 150^{\circ} - 20^{\circ} = 130^{\circ}$.$\angle BPD = 2(50^{\circ}) + 30^{\circ} = 100^{\circ} + 30^{\circ} = 130^{\circ}$.Thus,$x = 50^{\circ}$ and both angles measure $130^{\circ}$.

$\angle APC$ and $\angle BPD$ are vertically opposite angles. If $\angle APC = 3x - 20^{\circ}$ and $\angle BPD = 2x + 30^{\circ}$,then find $x$ and the measure of both angles.

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