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14 Jul 2021, 01:30
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B
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86% (00:53) correct
14%
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If AD = DB, DC = CB, and angle ABC = 90º as shown above, then what is the value of xº?
A. 40º
B. 50º
C. 65º
D. 80º
E. 100º
Attachment:
1.png [ 6.31 KiB | Viewed 2440 times ]
14 Jul 2021, 02:23
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Bunuel
In isos triangle BDC:
2x + 100 = 180
2x = 80
x = 40
So, DBC = 90 - 40 = 50
x = 50(Why? triangle BAD is isos)
IMO, (B)!
14 Jul 2021, 02:33
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Explanation:
In Triangle DBC,
DC=CB, Angle CDB = Angle DBC =y
Angle C = 100
Angle CDB + Angle DBC + Angle C = 180
2y + 100 = 180
y = 40
We know, Angle ABC is 90
Angle DBA = 90-40 = 50
Angle DBA = Angle DAB = 50
IMO-B
Posted from my mobile device
In Triangle DBC,
DC=CB, Angle CDB = Angle DBC =y
Angle C = 100
Angle CDB + Angle DBC + Angle C = 180
2y + 100 = 180
y = 40
We know, Angle ABC is 90
Angle DBA = 90-40 = 50
Angle DBA = Angle DAB = 50
IMO-B
Posted from my mobile device
