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14 Oct 2019, 00:31
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[GMAT math practice question]
The figure shows that line \(m\) is parallel to the line \(n\), and \(l\) is parallel to \(k\). Moreover, \(∠CBD=45^o, ∠FAE=80^o.\) What is \(∠BDC\)?
10.14ps.png [ 19.74 KiB | Viewed 1947 times ]
\(A. 40^o\)
\(B. 45^o\)
\(C. 50^o\)
\(D. 55^o\)
\(E. 60^o\)
The figure shows that line \(m\) is parallel to the line \(n\), and \(l\) is parallel to \(k\). Moreover, \(∠CBD=45^o, ∠FAE=80^o.\) What is \(∠BDC\)?
Attachment:
10.14ps.png [ 19.74 KiB | Viewed 1947 times ]
\(A. 40^o\)
\(B. 45^o\)
\(C. 50^o\)
\(D. 55^o\)
\(E. 60^o\)
14 Oct 2019, 01:30
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Angle at BAD =80°
Since ABCD is a parallelogram
Opp angles are equal and sum of angles is 360°
So angle ADC will be 100°
Angle CBD = Angle ADB =45°
So angle BDC will be 100-45= 55°
OA:D
Posted from my mobile device
Since ABCD is a parallelogram
Opp angles are equal and sum of angles is 360°
So angle ADC will be 100°
Angle CBD = Angle ADB =45°
So angle BDC will be 100-45= 55°
OA:D
Posted from my mobile device
14 Oct 2019, 20:40
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As m and n & k and l are parallel, ABCD is a parallelogram.
As ABCD is //, it's two opposite angles are same.
<EAF =80°, then<DAB is also 80°, as both are vertically opposite of the two straight lines.
It implies that <DCB =80°.
Triangle DCB, CBD+BCD+BDC= 180
80+45+BDC=180
BDC =55.
Hence, D.
As ABCD is //, it's two opposite angles are same.
<EAF =80°, then<DAB is also 80°, as both are vertically opposite of the two straight lines.
It implies that <DCB =80°.
Triangle DCB, CBD+BCD+BDC= 180
80+45+BDC=180
BDC =55.
Hence, D.
