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DBC + BCD or 90
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Updated on: 28 Nov 2018, 06:28
Updated on: 28 Nov 2018, 06:28
Expert Reply
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Question Stats:
87% (00:47) correct
12% (00:01) wrong based on 16 sessions
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triangle.jpg [ 9.24 KiB | Viewed 1931 times ]
\(AD=BD=6\), \(\angle ADB = 90°\)
Quantity A |
Quantity B |
\(\angle DBC+ \angle BCD\) |
\(90°\) |
A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
Kudos for R.A.E
Re: DBC + BCD or 90
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03 Nov 2018, 18:12
03 Nov 2018, 18:12
1
Carcass wrote:
Attachment:
triangle.jpg
\(AB=BD=6\), \(\angle ABD = 90°\)
Quantity A |
Quantity B |
\(\angle DBC+ \angle BCD\) |
\(90°\) |
A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
Kudos for R.A.E
Since \(\angle ADB = 90°\), therefore\(\angle BDC = 90°\)
Now in \(\triangle DBC\), we have
\(\angle BDC = 90°\)
and as the sum of angles in a \(\triangle\)add up to 180°
therefore
\(\angle DBC+ \angle BCD = 90°\)
Hence option C
Re: DBC + BCD or 90
[#permalink]
03 Nov 2018, 22:18
03 Nov 2018, 22:18
Expert Reply
Carcass wrote:
Attachment:
triangle.jpg
\(AB=BD=6\), \(\angle ABD = 90°\)
Quantity A |
Quantity B |
\(\angle DBC+ \angle BCD\) |
\(90°\) |
A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
Kudos for R.A.E
Ofcourse there is a typo as \(\angle ABD = 90°\) should be \(\angle ADB = 90°\)
now by the property of an exterior angle in a triangle, it is equal to the sum of opposite angles, so
\(\angle ADB = 90°=\angle DBC+ \angle BCD\)=A=90=B
C
Actually irrespective of AD=BD or not the answer will be C.
A poor question with wrong equivalence etc
Pl relook into the source.
