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In the figure above, if AD is parallel to BC, then ∠ADC =
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28 Apr 2020, 10:08
28 Apr 2020, 10:08
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#greprepclub In the figure above, if AD is parallel to BC, then.jpg [ 21.84 KiB | Viewed 2117 times ]
In the figure above, if AD is parallel to BC, then \(∠ADC = \)
A. 11°
B. 22°
C. 33°
D. 46°
E. 134°
Kudos for the right answer and explanation
Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
Given Kudos: 136
Re: In the figure above, if AD is parallel to BC, then ∠ADC =
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15 Jan 2022, 12:19
15 Jan 2022, 12:19
1
Carcass wrote:
Attachment:
#greprepclub In the figure above, if AD is parallel to BC, then.jpg
In the figure above, if AD is parallel to BC, then \(∠ADC = \)
A. 11°
B. 22°
C. 33°
D. 46°
E. 134°
Kudos for the right answer and explanation
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)°
Simplify this measurement to get (136 - x)°
Finally, since AD is parallel to BC, we know that the two highlighted angles below must add to 180°.
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
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Re: In the figure above, if AD is parallel to BC, then ∠ADC =
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07 Jun 2020, 04:16
07 Jun 2020, 04:16
1
AD is parallel to BC
So, Angle ADC + Angle BCD = 180°
Now, Angle BCD = Angle BCE + 2x
[Refer attached Image]
= 136°-x + 2x = 136° + x
Angle ADC + Angle BCD = 180°
=> 3x + 136° + x = 180
=> 4x = 180°-136° = 44°
=> x = 11°
So, Angle ADC = 3x = 3*11° = 33°
So, answer will be C
Hope it helps!
So, Angle ADC + Angle BCD = 180°
Now, Angle BCD = Angle BCE + 2x
[Refer attached Image]
= 136°-x + 2x = 136° + x
Angle ADC + Angle BCD = 180°
=> 3x + 136° + x = 180
=> 4x = 180°-136° = 44°
=> x = 11°
So, Angle ADC = 3x = 3*11° = 33°
So, answer will be C
Hope it helps!
Attachments
image-5.jpg [ 41.13 KiB | Viewed 1957 times ]
