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23 Jul 2020, 05:28
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
35%
(medium)
Question Stats:
71% (02:09) correct
29%
(02:35)
wrong
based on 170
sessions
History
Date
Time
Result
Not Attempted Yet
In the figure above, if AD is parallel to BC, then \(∠ADC = \)
A. 11°
B. 22°
C. 33°
D. 46°
E. 134°
Attachment:
1.jpg [ 21.84 KiB | Viewed 6367 times ]
23 Jul 2020, 05:42
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In Triangle BEC, \(\angle BCE\) = 180 - x - 44 = 136 - x (Sum of angles of a triangle = \(180^o\))
Since AD is parallel BC, \(\angle ADC\) + \(\angle BCD\) = \(180^o\) (Sum of interior angles between 2 parallel lines cut by a transversal = \(180^o\))
3x + 2x + 136 - x = 180
4x = 44
x = 11
\(\angle ADC\) = 3x = 3*11 = \(33^o\)
Option C
Arun Kumar
Since AD is parallel BC, \(\angle ADC\) + \(\angle BCD\) = \(180^o\) (Sum of interior angles between 2 parallel lines cut by a transversal = \(180^o\))
3x + 2x + 136 - x = 180
4x = 44
x = 11
\(\angle ADC\) = 3x = 3*11 = \(33^o\)
Option C
Arun Kumar
23 Jul 2020, 05:47
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Bunuel
Angle ADC = 3x = ?
Angle BCE = 180 - (X+44) = 136-x
Also, Angle BCD + Angle CDA = 180o (Parallel lines)
i.e. (136-x)+2x+3x = 180
i.e. 4x = 44
i.e. x = 11
hence, Angle ADC = 3x = 3*11 = 33o
Answer: Option C
