A
B
C
D
E
We can let x = the measure of angle C; thus, 2x = the measure of angle D, and x + 40 = the measure of angle B. We can create the equation:Vincen wrote: ↑Thu May 07, 2020 11:16 ampolygons_img5.png
In the figure above, in irregular quadrilateral \(ABCD,\) \(\angle A = 60^{\circ}, \angle D\) is twice \(\angle C,\) and \(\angle B\) is \(40^{\circ}\) more than \(\angle C.\) Find the measure of \(\angle B.\)
A. \(65^{\circ}\)
B. \(85^{\circ}\)
C. \(105^{\circ}\)
D. \(120^{\circ}\)
E. \(135^{\circ}\)
[spoiler]OA=C[/spoiler]
Source: Magoosh
Let, C = xVincen wrote: ↑Thu May 07, 2020 11:16 ampolygons_img5.png
In the figure above, in irregular quadrilateral \(ABCD,\) \(\angle A = 60^{\circ}, \angle D\) is twice \(\angle C,\) and \(\angle B\) is \(40^{\circ}\) more than \(\angle C.\) Find the measure of \(\angle B.\)
A. \(65^{\circ}\)
B. \(85^{\circ}\)
C. \(105^{\circ}\)
D. \(120^{\circ}\)
E. \(135^{\circ}\)
[spoiler]OA=C[/spoiler]
Source: Magoosh
