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In the figure above, if \(AD\) is parallel to \(BC,\) then \(\angle ADC=\angle ADC=\)

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In the figure above, if \(AD\) is parallel to \(BC,\) then \(\angle ADC=\angle ADC=\)

by M7MBA » Sun Nov 28, 2021 1:16 am

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In the figure above, if \(AD\) is parallel to \(BC,\) then \(\angle ADC=\angle ADC=\)

A. \(11^{\circ}\)

B. \(22^{\circ}\)

C. \(33^{\circ}\)

D. \(46^{\circ}\)

E. \(134^{\circ}\)

Answer: C

Source: Princeton Review
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Re: In the figure above, if \(AD\) is parallel to \(BC,\) then \(\angle ADC=\angle ADC=\)

by Brent@GMATPrepNow » Sun Nov 28, 2021 6:47 am
M7MBA wrote: ↑
Sun Nov 28, 2021 1:16 am
 1.jpg

In the figure above, if \(AD\) is parallel to \(BC,\) then \(\angle ADC=\angle ADC=\)

A. \(11^{\circ}\)

B. \(22^{\circ}\)

C. \(33^{\circ}\)

D. \(46^{\circ}\)

E. \(134^{\circ}\)

Answer: C

Source: Princeton Review
First, since angles in a triangle must add to 180°, we can see that the missing angle in the red triangle must be 180° - (x + 44)°
Image


Simplify this measurement to get (136 - x)°
Image


Finally, since AD is parallel to BC, we know that the two highlighted angles below must add to 180°.
Image

So, we can write: (136 - x)° + 2x° + 3x° = 180°
Simplify: 136 + 4x = 180
Subtract 136 from both sides: : 4x = 44
Solve: x = 11

Our goal is to find the measurement of ∠ADC
Since ∠ADC = 3x°, we can replace x with 11 to get: ∠ADC = 3x° = 3(11)° = 33°

Answer: C
Brent Hanneson - Creator of GMATPrepNow.com
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