In the figure above, in irregular quadrilateral $ABCD,$ $\angle A = 60^{\circ}, \angle D$ is twice $\angle C,$ and

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Vincen: Legendary Member; Posts: 2898; Joined: Thu Sep 07, 2017 2:49 pmFollowed by: 5 members; Thanked: 6 times

by Vincen » Thu May 07, 2020 11:16 am

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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

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Source: Beat The GMAT — Problem Solving | Thu May 07, 2020 11:16 am

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Re: In the figure above, in irregular quadrilateral $ABCD,$ $\angle A = 60^{\circ}, \angle D$ is twice $\angle C,$

by Scott@TargetTestPrep » Sat May 09, 2020 12:02 pm

Vincen wrote: ↑
Thu May 07, 2020 11:16 am
polygons_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

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:

∠A + ∠B + ∠C + ∠D = 360

60 + x + 40 + x + 2x = 360

4x = 260

x = 65

Therefore, the measure of angle B is 65 + 40 = 105 degrees.

Answer: C

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Re: In the figure above, in irregular quadrilateral $ABCD,$ $\angle A = 60^{\circ}, \angle D$ is twice $\angle C,$

by GMATinsight » Sat May 09, 2020 11:22 pm

Vincen wrote: ↑
Thu May 07, 2020 11:16 am
polygons_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 = x
then B = x+40
and D = 2x

Sum of angles in any quadrilateral = 360

A+B+C+D = 360
60+(x+40)+x+2x = 360
i.e. 4x = 260
i.e. x = 65

Now, B = x+40 = 65+40 = 105

Answer: Option C

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